School Science Lessons
(UNPh15)
2024-07-25
Circular motion, Pendulums and oscillations
Contents
15.0.0 Circular motion and vibration
15.3.0 Angular acceleration
15.5.0 Rotation, central forces
15.2.0 Centripetal force and centrifugal motion
15.4.0 Motion in two dimensions
15.4.3 Motion of centre of mass
15.1.0 Pendulums and oscillations
15.7.0 Projectiles
15.8.0 Relative motion
15.4.2 Velocity, position, acceleration
15.3.0 Angular acceleration
15.3.0.0 Angular acceleration
6.3.3.16 Radian, Angular velocity (GIF)
15.3.0.2 Rotational and translational kinetic energy of the earth
15.3.0.1 Rotational kinetic energy
15.3.0.3 Sliding and rolling
15.3.0.5 Spinning ice skater, angular momentum
15.3.0.4 Torque and angular acceleration
15.5.0 Rotation, central forces
Rotary Motion Accessory Kit, (Scientrific), Commercial
Experiments
15.5.5 Balls in water centrifuge
15.5.2 Empty jug by swirling
15.5.1 Flattening earth
15.5.6 Inertial pressure gradient
15.5.7 Turning motor car, with candle and balloon
15.5.13 Rotate a ball on a string
15.5.9 Rotate a candle
15.5.8 Rotate a doll
15.5.4 Rotate a manometer
15.5.12 Rotate a rubber wheel
15.5.11 Paper saw
15.5.3 Water parabola
15.2.0 Centripetal force and centrifugal motion
15.2.01 Centripetal force and centrifugal motion
Experiments
15.5.13 Ball on a string
15.2.16 Ball in a megaphone
15.2.12 Balls on a propeller
15.2.15 Banked track
15.2.21 Candle flame on a turntable
15.2.8 Carnival ride model, "wall of death"
15.2.5 Centrifuge, milk separator
15.2.4 Centrifugal force, clothes dryer, swinging bucket
15.2.2 Centripetal force, frequency and radius of uniform circular motion
4.161 Measure centripetal force
4.158 Centripetal force with a liquid
15.2.3 Centripetal force with a liquid in plastic goldfish bowl
4.160 Centripetal force with a swinging water bucket
15.2.1 Coin on a coat hanger, revolving coin
15.2.11 Conical pendulum
15.6.2 Crossing the river
15.6.1 Disc circle with gap, broken ring
15.6.4 Grinding wheel
15.2.14 Hand rotator
15.2 19 Lift with spin
6.3.3.16 Radian, Angular velocity
15.2.18 Raise a ball without touching
15.6.3 Release ball on a string
15.2.22 Rising napkin ring
15.2.17 Rolling chain
15.6.5 Spinning disc
15.2.9 Swinging up a weight
15.2.3.2 Torsional oscillations of a suspended rod
15.2.5.1 Turning a glass
15.2 20 Turning the corner
15.2.13 Welch centripetal force, spring stretching
15.2.7 Whirligig
15.0.0 Circular motion and vibration
15.0.2 Amplitude, A and period, T
15.0.4 Angular velocity, ω, radians, rad
15.0.4.1 Axis of rotation
15.0.5 Centripetal acceleration, ac
15.0.6 Centripetal force, Fc
15.0.1 Circular motion velocity
15.0.3 Frequency, f, simple harmonic motion
Experiment
15.0.1.1 Circle from straight lines
15.4.0 Motion in two dimensions
Experiments
15.4.1.2 Ball in a tube
15.4.1.3 Cycloid generator
15.4.2.2 Kick a moving ball
15.4.2.4 Passing the train
15.4.2.7 Sliding weights on triangle
15.4.1.4 Spots on a globe
15.4.3 Motion of the centre of mass
Experiments
15.4.3.5 Air table centre of mass
15.4.3.4 Centre of mass disc
15.4.3.6 Earth moon system
15.4.3.1 Lifting
15.4.3.3 Spinning, dry ice puck
15.4.3.2 Throwing
15.1.0 Pendulums, oscillations
15.2.11 Conical pendulum
15.1.3.1 Coupled oscillators
15.1.3 Coupled pendulums
15.1.5 Energy transfer between pendulums, by resonance
15.1.30 Foucault pendulum
15.1.29 Galileo's pendulum, stopped pendulum, conservation of mechanical energy
15.1.01 Pendulum, oscillations
15.1.02 Pendulum swinging at the equator
15.1.1 Simple pendulum
15.1.2 Simple pendulum, prove relationship of T to L
15.1.4 Stopped pendulum
15.7.0 Projectile motion, (Experiments)
15.7.01 Projectile horizontal and angular projection
A fired artillery shell travels througha parabolic path. The maximum range, by calculation, is when the shell is fired at an angle of 45o.
However, such a calculation ignores air resistance.
Air resistance may depend on other variables, e.g. the the shape of the projectile, the weight of the projectile, the direction and strength of the wind.
However, air resistanc may hardly affect the maximim range of an aerodynamically-shaped artillery projectile, but the angle to achieve actual maxim distance only be determined by practice shots. Experiment The water stream from a hose breaks up into droplets on contact wit the air.
Find the angle for maximum distance for water from a garden hose, when the water supply is partly turned on, about half turned on, fully tuned on.
Experiments
15.7.3 Midair target
15.7.5 Parabolic trajectory
15.7.1 Projectile cart, ballistics cart, gun and tunnel, howitzer and tunnel
15.7.4 Range of a gun
15.8.0 Relative motion
Moving reference frames
15.8.1.1 Crossing the river
Rotating reference frames
15.8.3.6 Chalk line on a turntable
15.8.3.11 Deflected water stream
15.8.3.7 Rolling ball on turntable
15.8.3.10 String through spinning globe
15.4.2 Velocity, position and acceleration
15.4.2.6 Brachistochrone track
15.4.2.6.1 Energy required to loop the loop
15.4.2.5 Galileo's circle
15.4.2.3 High road low road
15.4.2.2 Kick a moving ball
15.4.2.4 Passing the train
15.4.2.7 Sliding weights on triangle
15.0.1 Circular motion velocity
Circular motion velocity, circular velocity, v, uniform circular motion in a horizontal plane ac = v2/ r = 4π2r/ T2.
A body moving in a circular path is accelerating.
So a resultant force must reacting on it, acting towards the centre of the circle, called the centripetal force.
F = ma, so F = mv2 / r, for an object attached to a string whirled in a horizontal circle, the centripetal force is the tension in the string.
Circular motion velocity is always at a tangent to the circle.
For circular motion at constant speed, the centripetal ("centre seeking") acceleration vector always points towards the centre of the circle.
The centripetal force keeps the object moving in the circle and the direction of the force is towards the centre of the circle.
Centripetal force comes from the following:
1. Gravitational force, for planets in orbit, the sun and satellites about the earth
2. Tension in a string, for an object whirled in a horizontal circle
3. Electrical force, for electrons orbiting a nucleus
4. Friction, for a car turning
5. Air pressure, for an aircraft turning in a horizontal circle.
15.0.2 Amplitude, A and period, T
An oscillation or cycle is one complete to and fro movement of a vibrating object.
The maximum displacement of the vibrating object from its rest or equilibrium position is called the amplitude of the oscillation, with symbol A.
For any particular vibration, the time for one complete oscillation is called its period, with symbol T.
15.0.3 Frequency, f, simple harmonic motion
The number of oscillations or cycles in one second is called its frequency, f.
The relationship between the period and frequency of an oscillating object: T = 1/ f.
An oscillation may be very complicated, including more than one frequency.
So the oscillation of interest may comprise other oscillations.
If the oscillation can be described by a harmonic function with a single frequency, it is called simple harmonic oscillation.
An object moving at a constant speed in a circular path is undergoing uniform circular motion.
Calculate the speed or linear velocity of the object by dividing the length of the object's path by the time taken to complete the path.
For a complete circle the time taken is called the period, T.
Linear velocity (tangential velocity), v = 2 π r/ T, where r = radius of the circle and T is the time period for one revolution with unit ms-1, and is always at a tangent to the circle.
Simple harmonic motion is symmetrical oscillatory motion under a retarding force proportional to the displacement from the equilibrium position, e.g. the projection of uniform motion in a circle, circular motion, on to a diameter.
15.0.4 Angular velocity, ω, radians, rad
See diagram 15.0.4.2: Angular velocity, ω (omega).
See diagram 15.0.4.0: Rigid body motion.
Rigid body motion
t = time, sec = seconds
rev = revolutions, rpm = revolutions per minute
θ (theta) = angle
π (pi) = 22/7
ω (omega) = angular velocity
α (alpha) = angular acceleration
Measure angle in radians, (rad)
θ (theta) = S/r = 2πr/r = 2π rad
360o = 2πrad, 1o = 0.01745 rad
1 rad = 57.3>sup>o
Angular velocity
ω = θ/t rad/sec
θ = S/r = vt/r
ω = θ/t = v/r
v = θ r
1 rpm = 1 rev/min X 2pi rad/rev X 1/60 sec/min = 0.105 rad /sec
An LP record spins at 331/3 X 0.105 = 3.5 rad/sec
1. For an object moving in a circle, its speed around the circle, v = 2π r / T, where r = radius of the circle and T = time taken for one revolution.
2. The angular displacement of the object, ∆θ, is measured in radians, rad.
A radian is the angle subtended at the centre of a circle by an arc equal to the radius of the circle.
1 revolution, rev = 360o = 2π radians, rad, so 1 rad = 360o / 2π = 57.272727' = 57.30o.
Similarly, 1o = 0.01745 rad and 1 revolution per minute, 1 rpm = 0.105 rad/ sec.
3. Angular velocity, ω, = ∆θ / ∆t, radians per second, rad / s, where ω = change in angular position of the object / time, ∆θ = angular displacement, and ∆t = time taken.
(So linear velocity = m / s, and angular velocity = rad / s.)
4. The tangential velocity of an object moving with uniform circular motion is the ratio of its linear speed and the radius of its path, so v = ωr, where ω is expressed in radians, and time for one revolution, T = 2π / ω.
5. A 12 inch (12 × 2.54 cm = 30.48 cm) record turntable, radius of revolution, gramophone record rotates at 331/3 revolutions per minute, rpm = 3.49 rad / sec.
6. If a CD, compact disc rotates at 22.0 rad /s its angular speed = 210 rpm.
7. The earth spins once in 24 hours, 86, 400 seconds, so angular velocity = 4.16 × 103 degrees per second, i.e. 7 × 10-5 rad / s.
See diagram 15.0.4.1: Axis of rotation of the Earth.
15.0.4.1 Axis of rotation
The axis of rotation of an object is a line of particles, or imaginary particles, that do not move.
All other particles in the body move in a circle, V = ω, so the further particles are from the axis of rotation the faster they move, although they all have the same angular velocity.
A point on the rim of a gramophone record rotating at 331 / 3 rpm moves with velocity, v = ω r = 3.50 rad/ sec × 0.5 × 30.48 = 53.34 cm/ sec.
15.0.5 Centripetal acceleration, ac
An object undergoing uniform circular motion is constantly changing direction, so it is accelerating.
If the object is moving with constant speed the direction of acceleration is towards the axis of rotation, i.e. centre of the circular path.
This acceleration is called centripetal acceleration, ac = v2/ r, where r = radius and v = tangential velocity around the circle = 4π2r/ T2.
15.0.6 Centripetal force, Fc
Newton's First Law states that an object at rest or at constant velocity will remain in that state unless acted upon by an external force.
An object in circular motion, at constant speed v, must have a force acting on it to keep it in a circular path.
This centripetal force can be expressed as a form of Newton's Second Law, F = ma, i.e. Fc = mv2 / r = m4π2r/ T2.
When you observe an object undergoing uniform circular motion from a position either above or below the object's path, you see the circular path.
However, when the object is observed from a position in the same plane as the circular path, you see the object moving back and forth in an oscillatory motion that corresponds with the simple harmonic motion.
15.0.1.1 Circle from straight lines
Make a cardboard rectangle 2 cm × 10 cm.
Put the rectangle on a piece of paper on the desk.
Push a drawing pin (thumb tack) through the centre of the cardboard rectangle.
Move the cardboard from place to place drawing a line along the end of the ruler at each place.
Draw a dot where the lines cross.
Join the dots to draw a circle
15.1.01 Pendulum, oscillations
A simple pendulum is a mechanical system that displays oscillatory motion.
It consists of a mass, called a bob, suspended by a light cord that does not stretch.
For angles < 20o, the period of oscillation is dependent only on the length of the cord.
T = 2 π l / g, where l is the length of the pendulum and g is gravitational acceleration.
The period is independent of the mass of the bob.
Pendulums are used as timekeepers and to measure the acceleration due to gravity.
Period = 1/ frequency, Hertz, Hz, or cycles per second, cps.
The period, T, of a simple pendulum depends on its length, l, and the acceleration of a falling object, g.
T = 2π (l / g).
The period, T, is not affected by the angular amplitude, or the mass of the pendulum bob, m.
The simple pendulum does not display simple harmonic motion precisely and therefore is dependent on the angular amplitude.
However, for angles less than 20o the dependence on angular amplitude can be ignored.
15.1.02 Pendulum swinging at the equator
A pendulum swings slower at the equator for two reasons:
1. A "centrifugal" force of rω2 from the Earth's rotation decreases the acceleration by 0.034 m /s2, but this does not occur at the north pole and south pole.
2. The Earth bulges at the equator, so any place there is further from the centre of the Earth.
Consequently, the earth's gravitational acceleration is 9.780 m / s2 at the equator, about 0.052 m / s2 less than at the north pole and south pole, i.e. 0.53% less gravitational acceleration.
15.1.1 Simple pendulum
See diagram 15.1.1: Pendulum calculation.
A ceiling hook can be used to support a pendulum.
1. Tie a 2 m string to a heavy object.
Swing the pendulum bob through a small angle, < 20o, between the pendulum and a vertical line at the bottom of the swing.
Count the number of swings per minute.
Each swing starts and finishes when the bob passes the rest position in the same direction.
Swing the pendulum bob through a smaller arc.
Count the number of swings per minute.
The size of the arc does not affect the time of vibration of a pendulum.
The angular amplitude does not affect the period if the pendulum bob swings through a small angle.
2. Use the same 2 m pendulum, but change the weight of the bob.
Count the number of swings per minute.
The change in weight does not affect the time of oscillation of the pendulum.
The mass of the pendulum bob does not affect the period.
3. Change the length of the pendulum to two metres. Count the number of swings per minute.
The number of swings per minute increases.
The length of the pendulum affects the time of oscillation of the pendulum.
4. Measure the time taken for 25 oscillations through the same small angle for pendulum length, l = 0.25, 0.50, 0.75 and 1.00 m.
Calculate the time for one swing, T, for each length.
Assume the square roots of length, l, are 0.5, 0.71, 0.87 and 1 m.
Draw a graph to plot T against length, l.
If the graph line is a straight line, then T is proportional to the square root of length l.
The period, T, is the time taken to complete on one complete oscillation, forwards and backwards.
Period, T = 1/ frequency, f (expressed in hertz, Hz, cycles per second, cps) (Heinrich Hertz 1857 - 1894).
The period, T, of a simple pendulum depends on its length, l, and the acceleration, because of gravity (acceleration of free fall)
g = 9.8 m/ sec2 T = 2 × π × (l / g).
If the period, T = 1 second, T = 2 × π × (l/ g) l = g × T2/ 4 × π2, so length, l = 0.25 m.
A simple pendulum, length 25 cm, swings through one complete cycle every second.
Simple harmonic motion refers to movement of equal distance each side of a central point with acceleration towards the central point and proportional to the distance to it.
The bob moves with simple harmonic motion, because the force on the bob at any point is proportional to its displacement from its mean position and is directed towards it.
5. Tie a two metre long string to a heavy object. Swing the pendulum bob through a maximum angle of 20o between the pendulum and a vertical line at the bottom of the swing.
Count the number of swings per minute.
* Swing the pendulum bob through a smaller arc.
Count the number of swings per minute.
Note that the length of the arc does not affect the time of vibration of a pendulum.
* Keep the length of the pendulum the same, but change the weight of the bob.
Count the number of swings per minute.
Note that the change in weight does not affect the time of oscillation of the pendulum.
* Make the pendulum one metre long.
Count the number of swings per minute.
The number of swings per minute increases.
The length of the pendulum does affect the time of oscillation of the pendulum.
6. To verify the relationship between period of a simple pendulum and 1. angle of the swing, 2. mass of the pendulum bob, and 3. length of the pendulum.
The period is the time for a complete cycle of motion, i.e. forwards then back to where it started from.
The angle of the swing is the angle between its farthest position from the vertical and the vertical.
The length is the distance from the centre of the spherical bob and the point of suspension.
7. Use about 2 metres of fishing line or thread. Tie one end to a heavy circular bob.
Wind the other end around a metal cylinder.
Clamp the cylinder horizontally.
Turn the cylinder so that the length of the pendulum is exactly 1 metre.
Draw back the bob 10o and set the pendulum in motion.
Time how long it takes for the pendulum to complete 10 oscillations.
Calculate and record the period.
Pull the bob 20o and calculate the period.
Let the pendulum swing in a smaller arc by drawing the bob back 5o and calculate the period.
Note that the period is the same in each case, thus period is not dependent on size of swing.
8. Exchange the bob with another of different mass and repeat the procedure.
Observe that the period is the same.
The period is not dependent on the mass of the bob.
Construct a pendulum 2m long.
Using a protractor, draw back the bob 15o, release the bob and calculate the period.
Now reduce the length of the pendulum by 50 cm and repeat the procedure and calculate the period.
Repeat this procedure reducing the length of the pendulum each time.
Calculate 2 π × l/ g.
Table 15.1.1
Mass of bob, m (kilograms, kg) Length of pendulum, l (metres, m) Angular amplitude (angle of swing), θ (oC) No.
cycles (complete swings)		 Time
(for 10 cycles) (seconds, s)		 Average time (Time taken/ 10) (seconds, s) Period T (seconds, s) (calculate: T = 2 π l/ g)
1 kg 1 metre 15oC 10 . . .
1.5 kg 1 m 15oC 10 . . .
1 kg 1.m 10oC* 10 . . .

15.1.2 Simple pendulum, find and prove the relationship of T to L
To make a simple pendulum, hang a weight from a soft cord more than 2 m length.
Draw back the object 15o then let it swing in a vertical plane.
Record the time for a complete swing, forwards then back.
Let the pendulum swing in a smaller arc and record the time for a complete swing.
Compare the times of the two angles of swing.
The length of the arc does not affect the time of the swing.
Change the weight of the object and let the pendulum swing again with the same angles and length of cord.
Record the times of swing.
The weight of the object does not affect the time of swing.
Change the length of the cord.
Repeat the experiment with the same angles of swing and the same weights.
The length of the pendulum affects the time of the swinging.
Make a form using the data.
List the length of the pendulum l in turn according to the length at the first line.
List the corresponding vibrating period T (T = 60/ the time of swinging per minute) at the second line.
You will find out the relationship between vibrating period T and the length of the pendulum, L.
15.1.3 Coupled pendulums
See diagram 15.1.3: Coupled pendulums.
1. Demonstrate the transfer of energy in a weakly coupled system.
Fill two same size bottles with water, add stoppers and suspend the bottles with same size string, as pendulums from a rod.
The strings must be the same length.
Hold one bottle still, start the other bottle swinging, then release the first bottle.
Soon the swinging pendulum slows, and the other pendulum takes up the swing.
Hang the pendulums from a stationary support such as a doorway, but join the pendulum cords with a third string tied between them about one eighth of the way down the cords.
2. To observe the energy transfer of a simple pendulum use two full identical bottles, several pieces of string, a stick, and two chairs.
Fill the two bottles with water or sand.
Put the chairs back to back.
Lay the stick across the backs of the chairs.
Cut two pieces of string with the same length.
Tie the bottlenecks with string then hang the bottles from the stick.
When the bottles are at rest, hold the first bottle and draw back the second bottle.
Let the second bottle swing in the plane upright to the stick.
Let the first bottle go, then observe the movement of the two bottles.
The amplitude of the second bottle will decrease gradually.
The first bottle will gradually swing from rest and its amplitude will increase gradually.
Finally the two bottles will swing with the same amplitude and velocity.
15.1.3.1 Coupled oscillators
See diagram 15.1.3.1: Coupled oscillators on an air track.
Use springs to connect air track gliders to the binding posts and to each other, and show the different modes of oscillation demonstrated.
A high frequency is realized when the air track gliders moving in opposition move with high frequency.
Air track gliders moving with parallel velocities move with low frequency.
15.1.4 Stopped pendulum
See diagram 15.1.4.0: Stopped pendulum.
See diagram 15.1.4.1: Stopped pendulum with metre stick.
1. Set a pendulum swinging and note the maximum height reached by the pendulum bob.
Place a stop that has the effect of reducing the length of the pendulum.
Set the pendulum swinging as before and the bob still returns to the height of the original motion.
2. Raise the pendulum a certain height.
It will travel to nearly the same height on the opposite side whether it is "stopped" or not.
The metre stick serves as a reference for the height of the pendulum.
15.1.5 Energy transfer between pendulums, by resonance
See diagram 15.4.12, Energy transfer between pendulums.
Study how the time taken for energy transfer between pendulums depends on 1. the distance between hanging points of the pendulums and 2. the length of the pendulums.
Suspend a 100 cm strong string between two stands.
Attach two threads 2.5 cm each side of the centre of the strong string.
Attach 100 g weights to the end of each thread so that the length of the thread is 50 cm.
Pull one weight to the side through a 60o angle to the vertical.
While noting the time in seconds, release the weight so that it swings freely back and forth as a pendulum, but does not touch the stationary second pendulum.
The energy of the first pendulum transfers to the second pendulum.
The first pendulum swings less until it stops swinging and the second pendulum swings more until it has the original swing of the first pendulum.
Note the time when the first pendulum stops.
The energy of the first pendulum transfers to the second pendulum.
Note the time when the second pendulum stops.
Note the times for five transformations of energy.
Calculate the average time needed for one transformation of energy.
Repeat the experiment by increasing the distance between the hanging points of the pendulums.
Repeat the experiment by shortening the length of the thread.
Repeat the experiment by changing the initial angle of swing.
How does time of transfer depend on the following:
1. distance between pendulums,
2. length of pendulums,
3. original angle of swing of pendulums?
Note that the distance between pendulums affects the tension in the strong string.
Table 2.2.2
Experiment distance
d		 length
l		 angle
a		 Transfer
(seconds)		 Total
(seconds)		 Average
(seconds)
1 (control) 4.5 50 60o . - -
4 (distance) 10 50 60o . . .
3 (length) 4.5 100 60o . . .
4 (angle) 4.5 50 30o . . .

15.1.29 Galileo's pendulum, stopped pendulum, conservation of mechanical energy
See diagram 15.1.20, Galileo's pendulum.
In about 1602, Galileo Galilei while observing the oscillation of a hanging lamp in the cathedral of Pisa and comparing it with his pulse, discovered the isochronism of the pendulum, i.e. the period of the swing is independent of the amplitude.
This principle became the standard mechanism for clocks.
If Galileo Galilei had the patience to sit in the cathedral of Pisa for 24 hours, he would have done the experiment of 15.1.30 Foucault pendulum!
Experiments
1. Suspend a 5 cm diameter wooden ball by a 100 cm fine thread from a nail in the wall or from a nail above the blackboard.
Draw a horizontal line CD 50 cm above the wooden ball such that the wooden ball can swing from C to D.
Release the wooden ball at C so that it swings through the arc CBD to reach D.
The momentum at B carries the wooden ball through the arc BD to the horizontal CD.
In fact the ball does not quite reach the horizontal CD, because some energy is lost due to resistance of the air and the string.
2. Place a peg as a stop at some position E between the nail and CD.
Release the wooden ball at C.
The ball again travels through arc CB, hits the peg E then travels though arc E E1 reaching the same height as CD.
3. Replace peg E with peg F between E and level CD.
Release the wooden ball at C.
The ball again travels through arc CB, hits peg F then travels through arc FF1 reaching the same height as CD.
4. Replace peg F with peg G such that the remainder of the thread below G cannot reach level CD.
Release the wooden ball at C.
The thread leaps over peg G and twists around it.
15.1.30 Foucault pendulum
See diagram 36.87: G-clamp support to allow pendulum to swing in any direction.
In 1851, Léon Foucault hung a heavy brass ball from a long wire attached to the inside of the dome of the Pantheon in Paris and set it in motion in one direction as a pendulum.
He showed that the earth rotates once each day, because while the direction of swing of the pendulum does not change the surroundings do change their position due to the rotation of the earth.
As the pendulum swings, the Earth rotates beneath it, so the plane of the swing appears to rotate.
Experiments
1. Use a G-clamp with a ball bearing soldered to the inside of the jaw to makes a good support for the pendulum.
Hang the pendulum indoors with the ball bearing resting on a razor blade or another hard surface.
Use nylon fishing line to suspend the bob.
The bob can be a solid rubber ball, but for best results use a bob > 5 kg.
For a pointer, use a short knitting needle pushed into the bob and continuous with the suspending fishing line.
The pointer should just touch a reference line drawn in fine sand in a tray on the floor.
The length of the pendulum at least 3 m long.
The pendulum must swing freely so that it swings backwards and forwards in the same plane.
However, it appears to change its path during the day.
2. To set the pendulum in motion, attach a long cotton thread to a drawing pin pushed into the bob.
Align the thread along the direction of the reference line, then burn the thread near the drawing pin.
After the pendulum is set in motion note that the plane of the swing has changed after a few hours compared with the reference line.
If the ceiling-mounted pendulum swings freely, note the change in the path of the pendulum after one hour.
Then note the plane of swing at six × ten minute intervals.
A pendulum releasing ink can mark a clear pattern.
Getting good quantitative results without many refinements is not easy, but observing the effect is not difficult.
3. Note the variation of rotation of the Foucault pendulum with latitude.
The Earth rotating beneath the bob causes the change.
The precession period for an ideal pendulum is 23.93 hours / sine of the latitude.
For example, at Sydney, Australia, at latitude 34o S, the period is about 43 hours, i.e. about one degree every seven minutes.
At the south pole the pendulum precesses through 360o in a day.
At the equator the pendulum does not precess.
4. If a pendulum is started swinging along the north-south direction, and below it a model of the Earth and the and pendulum is slowly rotated in a counter clockwise direction.
The pendulum continues its swing along the North-South direction until after the earth model has been rotated by 90o.
On the equator, the pendulum does not precess.
At intermediate latitudes the period = 23.93 hours / sine of the latitude, e.g. at Sydney's latitude of 34 degrees S, the period = about 43 hours, so the precession rate = about 1o every 7 minutes.
It is easy to deal with this misunderstanding.
The plane of a pendulum at the equator, swinging in a North South plane, does not rotate with respect to the earth.
Relative to an inertial frame, it rotates once every 24 hours.
Only at the north and south poles, does the pendulum swing in a plane fixed with respect to the distant stars while the Earth rotates beneath it, but not at all other latitudes, where the plane of the pendulum's motion rotates with respect to an inertial frame.
At these other latitudes, the plane of the pendulum's swing is not fixed in space.
5. Construct a miniature Foucault pendulum
Mount a small Foucault pendulum from a stand set upon a turntable or office chair that can be rotated.
Observe the behaviour of the pendulum when the turntable is rotated slowly.
Note that the plane of motion of the simple Foucault pendulum remains stationary as the support is rotated.
Set the pendulum into oscillations, then note and mark its plane of oscillation.
While the pendulum is still swinging, rotate its support frame.
The pendulum continues to oscillate in the original plane.
This demonstration simulates a pendulum oscillating above the poles of the Earth.
15.2.01 Centripetal force and centrifugal motion
When an object is travelling in uniform circular motion at a constant speed it is still accelerating.
The direction of the velocity changes, so there is an acceleration, called centripetal acceleration.
Centripetal force keeps an object in a circular path and changes the direction of the velocity towards the centre of the circle.
Centripetal force, F = mv2 / r, m = mass of the object, r = radius of the circle, v = linear, i.e. tangential velocity.
The centripetal force keeping the earth in its orbit around the sun is the gravitational force.
The centripetal force keeping an object sitting on a rotating turntable is friction.
If you stand on a rotating platform you feel as if you are being forced in a direction away from the centre of the circle.
This is called "centrifugal force", but it is the result of inertia.
Your body would like to travel in a straight line according to Newton's First Law.
As the platform floor accelerates towards the centre and you try to continue on your straight line motion, it feels as if a force pushes you to the outside.
However, there is no force.
Centrifugal force is a "fictitious" force that is observed from within accelerating frames of reference.
Of course as centripetal force disappears, the object will move along a path that is tangent of the circle at the point where the centripetal force vanished.
A simple pendulum consists of a cord whose mass can be ignored and a ball whose volume is very small called a "bob".
When the swinging angle is less than 5o, the simple pendulum moves with a constant period, T.
T is not affected by the mass of the bob.
T = 2π × ( l/ g) where l = length of the cord and g = acceleration of gravity.
4.158 Centripetal force with a liquid
See diagram 15.243: Centripetal force with a liquid.
Tie a wire securely around the neck of a plastic goldfish bowl.
Attach a string to the wire.
Clamp a hook in a hand drill chuck and attach it in the centre of the string.
Put water coloured with ink in the bowl.
Put a plastic "fish" in the bowl.
Turn the drill handle to spin the bowl and water or whirl the bowl around you by hand.
The water begins to climb up the sides of the bowl, because of the "centrifugal force" on the water, not the centripetal force on the water.
Note the effects of inertia of the water when starting and stopping spinning the bowl.
When starting spinning, the water tends to remain stationary.
When stopping spinning, the water tends to continue spinning inside the bowl.
4.160 Centripetal force with a swinging water bucket
1. Almost fill a small bucket with water.
Swing it around rapidly at arm's length in a vertical circle.
The water does not spill, because centrifugal, not centripetal, force acts on the water to keep it in the bucket.
2. Half fill a bucket with water.
Swing the bucket forwards and backwards.
The water "climbs up" the side of the bucket due to its inertia.
3. Some people can do the water in glass trick by turning a glass of water quickly through 360o, without spilling a drop.
4. Grasp a plastic cup full of water with your palm facing away from you.
Keep you arm straight while you swing the plastic cup in a smooth circle.
If you jerk your arm or stop the swing the water spills out.
Keep swinging your arm, but slow the swing until you stop the swing and place the plastic cup on the table with no water spilled out.
5. Swing a lighted lantern forwards and backwards the flame pointed to the direction of movement, because the heavier air around the lighter flame is left behind due to its inertia.
4.161 Measure centripetal force
See diagram 4.161: Measure centripetal force.
Use a metal or wooden tube about 15 cm long.
Tie a two-holes rubber stopper to the end of 1.5 m of string.
Pass the other end of the string through the tube and attach iron washers for weights.
Adjust the string so that the distance from the top of the tube to the cork is 1 m.
Grip the tube as a handle and swing it in a small circle above your head so that the rubber stopper moves in a horizontal circle.
The force of gravity on the washers provides the horizontal force needed to keep the stopper moving in a circle.
Attach a clip to the string below the tube.
Swing the tube such that the level of the clip remains constant and so the circular motion is constant.
To find the frequency, record the number of revolutions per minute.
Record the number of washers when the stopper moves in a path of radius 1 m at constant revolutions per unit time.
If you increase the number of washers, you must increase the speed of the stopper to keep the washers at the same height, F = mg = m (v2 / r).
If you halve the radius of the stopper, you must decrease the speed of the stopper to keep the washers at the same height, F = mg = m (v2/ r).
15.2.1 Coin on a coat hanger, revolving coin
See diagram 15.2.1: Whirling coat hanger.
Balance a coin on the hook of a coat hanger.
Balance the coin on the middle of the crossbar.
Hook the hanger over your finger.
Start swinging the hanger gently back and forth.
Twirl the coat hanger around your finger and the coin does not fly off.
OR
Pull the middle of the longest straight part of a wire coat hanger out and shape it into a long narrow shape.
Use a file to flatten the end of the hook.
Let the hanger hang on your index finger and place a coin on the filed end of the wire.
Start to swing the hanger.
Swing slowly at first back and forth, and then do full loops.
Swing it slower when ending the swing and try to catch the coin when it falls off the end of the hanger.
The spinning of the penny gives it a "centrifugal force" holding the coin against the end of the hanger.
The faster the spinning, the larger this centrifugal force; the slower the spinning, the smaller the force.
The force keeps the coin up against the hanger until the circular motion is too slow.
Centrifugal forces are applied in the automatic cloth washer, where the drum filled with clothes and water spins fast to separate the water from the clothes.
The tendency to stay moving in a straight line causes this force of inertia or centrifugal force.
[This experiment may need some practice before demonstrating to a class].
A revolving clothes dryer works in a similar fashion.
15.2.2 Centripetal force, frequency and radius of uniform circular motion
See diagram 15.2.2: Whirling stopper.
1. Research the relationship among the centripetal force, frequency and radius of uniform circular motion.
Use a 15 cm length of glass, inner diameter 1 cm.
Burn its end over Bunsen burner into round and smooth.
Twine two layers plastic adhesive tape or list on the surface of the glass to prevent the glass slip from your hand.
Use a piece of nylon fish line.
Use one end of the fishing line to tie a rubber stopper with two holes.
Let the fishing line go through the two holes and make a tie at the centre of the stopper.
Let the other end of the fishing line go through the glass then hang 6 iron gaskets, outer diameter 1 cm or 2 nuts, M10, and finally tie the line in a knot.
Let a clip go through the last tie to prevent the gaskets slipping away.
Adjust the length of the line, from the rubber stopper to the top of the glass, to 1 m.
Now hold the glass and on its top swing the rubber stopper in a level circle.
Here centripetal force keeping the stopper moving in a circular motion is the line's strain.
The force is caused by the weight of gaskets or nuts.
You may need to practice several times before demonstrating this experiment.
Record the amount of the original gaskets and the frequency of the uniform rotation.
Change the length of the level line so that the rotating radius goes up 0.5 times of the original size at first then down 0.5 times.
Differently count the time of the rotations per minute then exchange into frequency and record.
You will find how the centripetal force affects the frequency of the rotation.
2. Use a metal or wooden tube about 15 cm long. Tie a two holed rubber stopper to the end of 1.5 m of string.
Pass the other end of the string through the tube and attach iron washers for weights.
Adjust the string so that the distance from the top of the tube to the cork is 1 m.
Grip the tube as a handle and swing it in a small circle above your head so that the rubber stopper moves in a horizontal circle.
The force of gravity on the washers provides the horizontal force needed to keep the stopper moving in a circle.
Attach a clip to the string below the tube.
Swing the tube such that the level of the clip remains constant and so the circular motion is constant.
To find the frequency, record the number of revolutions per minute.
Record the number of washers when the stopper moves in a path of radius 1 m at constant revolutions per unit time.
If you increase the number of washers, you must increase the speed of the stopper to keep the washers at the same height, F = mg = m (v2/ r)
If you halve the radius of the stopper, you must decrease the speed of the stopper to keep the washers at the same height, F = mg = m (v2 / r).
15.2.3 Centripetal force with a liquid in a plastic goldfish bowl
See diagram 15.2.3: Whirling fish bowl.
1. Tie a wire securely around the neck of a plastic goldfish bowl.
Attach a string to the wire.
Clamp a hook in a hand drill chuck and attach it in the centre of the string.
Put water coloured with ink in the bowl. Put a plastic "fish" in the bowl.
Turn the drill handle to spin the bowl and water or whirl the bowl around you by hand.
The water begins to climb up the sides of the bowl, because of the "centrifugal force" on the water, not the centripetal force on the water.
Note the effects of inertia of the water when starting and stopping spinning the bowl.
When starting spinning, the water tends to remain stationary.
When stopping spinning, the water tends to continue spinning inside the bowl.
2. Observe liquid's circular motion and related phenomena.
Use a small plastic goldfish bowl or the lower part of a plastic drink bottle with two holes drilled each side of the rim.
Fasten a 1 m piece of string to the rim then hang it on the end of a spoon.
Fill it to 1/ 3 with water containing tea and tea leaves.
Hold the jar to make it immovable and twist the spoon in a level plane to make the strings twist together.
Raise the spoon to raise the bowl above the ground or table then take your hand off the jar.
Observe the rotation of the bowl and the rotation of the tea and the tea leave in the bowl.
The tea rises close to the side of the bowl.
Do the experiment again after the water is rest completely.
This time mostly observe the water's movement when the jar just begins to rotate and just stops.
Do the experiment once again.
When the jar rotates fastest brake it with your hands.
Observe the water's movement this time.
15.2.3.2 Torsional oscillations of a suspended rod
Suspend a rod by two strings attached symmetrically its centre of mass.
Displace the rod small distances in opposite directions so that the rod undergoes torsional oscillations.
Observe how the period of oscillation depend on the 1. Length of the strings, 2. Distance between the two strings, 3. Length of the rod, 4. Mass of the rod,
5. Angle of twist of the rod, 6. Number of twist rotations, i.e. observe whether the time taken to unwind the turns depends on the number of turns.
Compare this observation with twists in a child's playground swing.
15.2.4 Centrifugal force, clothes dryer
See diagram 15.2.4: Whirling clothes dryer.
Make a clothes dryer.
Use an empty container such as a can without lid or the bottom half of a plastic drink bottle cut horizontally.
Punch several holes around the container, then punch three holes equidistant from each other around the top of the container to suspend it with a thread.
Place wet clothes in the container and make sure that the whole container is in a state of even mass distribution.
Spin the container slowly in one direction to make thread gradually turn around.
Remove your hand suddenly, the container can spin rapidly and you will find that water is thrown from the container.
As the container spins rapidly, the adhesive force between water drop and clothes is smaller than centripetal force that water drop needed to move in a circular line.
Then water drop leave the clothes to do centrifugal motion.
As a result, it is thrown out from the clothes.
The household dryers are made according to the principle.
Note what relationships between the speed of spinning of the container and the amount of the water.
To strengthen the effect of the experiment, you may add more water.
15.2.5 Centrifuge, milk separator
See: Centrifuge (Commercial).
See diagram 15.2.5: Whirling pail.
1. Almost fill a small bucket with water.
Swing it around rapidly at arm's length in a vertical circle just fast enough so that the water doesn't spill out.
The water does not spill, because centrifugal, not centripetal, force acts on the water to keep it in the bucket.
2. Half fill a bucket with water.
Swing the bucket forwards and backwards.
The water "climbs up" the side of the bucket due to its inertia.
3. Half fill a plastic bucket with water.
Hold it by the handle and swing the bucket in a vertical circle.
If you swing fast enough the water does not spill out, because the water is pressing against the bottom of the bucket.
Keep swing and then let go the handle.
The bucket and water moves at a tangent to the direction of spin.
This action is the same as throwing a stone with a sling shot or throwing a lasso to land over the head of a runaway cow.
4. Swing a bucket of nails in a vertical circle over your head.
Swing a bucket of nails in a tin bucket.
The nails stay in the bucket, but you can hear the bottom layers of nails dropping away from the bottom of the bucket at the top of the swing, if swung in just the right way.
5. Use a hand cranked centrifuge to remove chalk dust from a solution.
Dissolve finely chalk dust in a test tube of water and pour the solution into one arm of a bench mounted centrifuge.
Fill the other arm with water to balance the solution.
Turn the handle of the centrifuge to force the chalk out of the solution.
6. A pilot, making a sharp turn feels himself pressed against the plane's cockpit.
7. In some fairgrounds people are spun around in a "wall of death".
The floor drops down, but the people remain pressed against the spinning wall.
6. A centrifuge spins a test-tube very high speed to separate substances according to their relative density.
8. A milk separator is used to separate the milk from the cream.
Similarly water can be separated from fuel.
9. Grasp a plastic cup full of water with your palm facing away from you.
Keep you arm straight while you swing the plastic cup in a smooth circle.
If you jerk your arm or stop the swing the water spills out.
Keep swinging your arm, but slow the swing until you gently stop the swing and place the plastic cup on the table with no water spilled out.
10. Swing a lighted lantern forwards and backwards the flame pointed to the direction of movement, because the heavier air around the lighter flame is left behind due to its inertia.
15.2.5.1 Turning a glass
Some people can do the water in glass trick by turning a glass of water quickly through 360o, without spilling a drop.
Put a glass full of water on the table.
Grab hold of the glass with palm of the hand facing away from you.
Then turn the glass quickly through 360o and put the glass back on the table without spilling a drop of water.
This demonstration requires a lot of practice and should be practised outdoors with a plastic cup.
15.2.7 Whirligig
Attach 1 kg and 100 g ball to the ends of a 1 m string passing through a hollow tube.
Twirl one ball then the other ball around your head.
15.2.8 Carnival ride model, "wall of death"
Toy person stands on the inside wall of a rotating cylinder.
15.2.9 Swinging up a weight
A swinging weight picks up a much heavier weight.
15.2.11 Conical pendulum
Use a ceiling-mounted bowling ball pendulum as a conical pendulum.
A model plane flies around on a string defining a conical pendulum.
The main bearing of a conical pendulum is from a bicycle front wheel axle where the string tension is set by a counterweight.
15.2.12 Balls on a propeller
Balls sit in cups mounted on a swinging arm at 0.5 and 1.0 m.
Calculate the period necessary to keep the ball in the outer cup and swing it around in time to a metronome.
Ride on a twirling device for children in a park.
15.2.13 Welch centripetal force, spring stretching:
Compare the angular velocity with the mass needed to stretch a spring a certain distance.
15.2.14 Hand rotator
Mount two 2000 g spring balances on a rotator.
Attach equal masses to each spring balance and read them at some rotational velocity.
15.2.14.1 Fijit spinner
Top toy spun in the hand, where user holds a centre pad and flicks one of three rounded blades to cause the spinner to rotate around a centre bearing.
It keeps spinning for a long time, because of its light weight and low friction around the centre bearing.
15.2.15 Banked track
Roll a steel ball down an incline into a funnel so that it reaches an equilibrium level where it revolves in a horizontal plane.
15.2.16 Ball in a megaphone
Throw a ball into a megaphone and it turns around and comes out the wide end.
15.2.17 Rolling chain:
Spin a loop of chain on and released to maintain rigidity and roll down the bench as a rigid hoop.
The spun flexible chain acts as a solid object
15.2.18 Raise a ball without touching
Put a marble or a small ball made of Plasticine (modelling clay), and an empty jar on a table.
Ask a friend to raise the ball without touching it.
Place the jar over the marble or ball.
Twist the jar very quickly so the ball is running around the entrance to the jar.
If you lift the jar the ball also rises while still spinning around the inside of the jar.
15.2.19 Lift with spin
Make two holes in each side of a plastic pot then join the holes with a string to make a handle.
Put marbles or stones in the plastic pot.
Tie one end of a 50 cm string to a weight.
Thread the other end through a ball pen case then tie it to the centre of the handle.
Twist the ball pen case so that the weight is spinning very fast.
Raise the pen case slowly.
The spinning weight will raise the pot of stones.
15.2.20 Turning the corner
Put a coin in the centre of a smooth wooden tray.
Move the tray forwards sharply.
The coin stays in the same place or moves forwards depending on the friction between the coin and the tray.
Move the tray forwards and to the right sharply.
The coin still moves forwards when the tray moves to the right and may hit the front of the tray.
The coin is like people in a motor car without seat belts when the car swerves sharply to the right to avoid hitting something.
15.2.21 Candle flame on a turntable
Mount a candle on a turntable, e.g. a record player.
Light the candle.
Put a large tall jar over the candle with space under the jar to allow air to come it and keep the candle alight.
Turn on the turntable.
The candle flame points towards the centre of the turntable.
Inside the jar the more dense air moves outwards, because of centripetal force so the less dense gases of the candle flame move inwards.
The centripetal force moves the more dense air more.
The flame with less mass than the air around it is accelerated more by the centripetal force.
15.2.22 Rising napkin ring
A napkin ring nowadays is just a ring of plastic.
Put the ring on the table and dip the pointing finger down to touch the inside of the ring.
Turn the finger in a circle so that the ring starts to spin around the finger.
15.3.0.0 Angular acceleration, α, tangential acceleration, αt, and centripetal acceleration, αc
See diagram 15.3.0: Angular acceleration.
If angular velocity changes from ω1 to ω2, angular acceleration, α = (ω2 - ω1) / time, t
v = ωr, so v1 = ω1 × r and v2 = ω2 × r, so α = (v2/ r - v1/ r)/ t = (v2 - v1)/ t × r
linear acceleration of a body tangent to its path, at = (v2- v1)/ t, so α = at/ r, and at = α × r, i.e. tangential acceleration = angular acceleration × radius.
Centripetal acceleration ac = v2/ r, towards the centre of the circle, so at and ac are at right angles
15.3.0.1 Rotational kinetic energy
See diagram: 15.3.0.1: Moments of inertia of bodies about axes.
Rotational kinetic energy
v = ω r, so for particle mass m, KE = mv2 = mω2r2, so for all particles in an object.
(All particles in the object have the same angular velocity.)
KE sum of mv2 = × (sum of m × r2) × ω2.
Moment of inertia of object, I = sum of m × r2 (Moment of inertia is a sort of rotational analogue of mass.)
KE of object, moment of inertia I, rotating with angular velocity ω, = Iω2.
In theory, for particles m1, m2 and m3 at radius r1, r2, r3 in an object, I = sum of m × r2 = m1 × r12 + m2 × r22 + m3 × r32.kg m2.
Moment of inertia is difficult to calculate for irregular bodies, however, for regular bodies about axes, it can be easily calculated.
15.3.0.2 Rotational and translational kinetic energy of the earth
See diagram: 15.3.0.1: Moments of inertia of bodies about axes.
Mass of the earth = 6 × 1024 kg.
Radius of earth = 6.4 × 106 metres.
If earth is a sphere, I = 2 / 5 MR2 = 2 / 5 (6 × 1024) × (6.4 × 106) = 9.8 × 1037 kg.m2.
Angular velocity of earth = 7.3 × 10-5 rad/ sec (one rotation per day).
Rotational kinetic energy, KE (Earth turns around its axis.) = Iω2 = × 9.8 × 1037 × 7.3 × 10-5 = 2.6 × 1029j.
Translational kinetic energy, KE
(Earth revolves around the sun at average orbital speed = 3 × 104 metres / sec.).
KE = × m × v2 = × 6 × 1024 × (3 × 104)2 = 2.7 × 1033j.
So the kinetic energy of orbital motion is much greater than the kinetic energy of rotational motion.
If mass of the earth = 6 × 1024 kg, and radius of earth = 6.6 × 106 metres, moment of inertia = 0.4 × M × R2 = about 1 × 1038 kg m2.
15.3.0.3 Sliding and rolling
A cylinder mass m and radius R is at the top of an inclined plane, PE = mgh
1. The cylinder slides down the slope with no friction.
It travels with velocity v at end of slope.
PE = KE, mgh = mv2 (translational kinetic energy at the bottom), so v = 2gh.
2. The cylinder rolls down the slope.
Moment of inertia of cylinder that rolls and does not slip = I = × m × R2. Angular velocity ω = v/ R, so ω2 = v2 / R2.
PE = KE, so mgh = mv2 (translational kinetic energy at the bottom) + × I × ω2.
(rotational kinetic energy at the bottom) = mv2 + ( mR2) (v2 / R2) = mv2 + mv2 = mv2, so v = 4 / 3 gh.
3. At the bottom of the inclined plane, the rolling cylinder moves slower than the sliding cylinder, because some energy is absorbed by rotation.
15.3.0.4 Torque and angular acceleration, line of action of a force, centre of gravity
See diagram 15.3.0: Torque.
A force f acts on an object mass m attached to the end of a string and moving in a circular path.
F = ma, and angular acceleration, α = a/ r, so F = m × r × α, and multiplying both sides of the equation by r: Fr = m × r2 × α.
Fr = torque = τ of the force F about the axis of the particle's rotation and its moment of inertia, I, = mr2.
So torque = I × α (analogous to F = Ma).
Angular acceleration, α is proportional to the torque and inversely proportional to the moment of inertia of the object.
When a force acts on an object such that the line of action of the force passes through the centre of gravity of the object, the object will accelerate according to f = ma.
However, if the line of action of the force is not through the centre of gravity, the object will have both linear and angular acceleration.
15.3.0.5 Spinning ice skater, angular momentum
Angular momentum, L = I × ω.
When there is no net torque on an object its angular velocity and angular momentum are constant.
Conservation of angular momentum:
When the sum of torques acting on an object = 0, the angular momentum is constant.
The greater the value of angular momentum, the more torque is needed to change its direction so bullets, rockets and even footballs can be aimed more accurate if spin stabilized.
Similarly, a spinning top stays upright until friction causes it to lose angular momentum.
An ice skater can start a spin on one toe with one leg extended and both arms extended (I is large and ω is small), but when the ice skater brings both legs and both arms together (now I is small and ω is large) the skater spins much faster due to conservation of angular momentum.
Angular momentum of the earth = moment of inertia × angular velocity = (1 × 1038) × (7 × 10-5) = 7 × 1033 newton metre (Nm).
15.4.1.2 Ball in a tube
Attach a ball to a string at the bottom of a vertical tube and hold the string while moving the tube horizontally.
15.4.1.3 Cycloid generator
Roll a hoop with a piece of chalk fastened to the circumference along the chalk tray.
Join large and small cylinders coaxially.
A spot on the larger cylinder moves in a cycloid when the smaller cylinder is rolled on its circumference.
15.4.1.4 Spots on a globe
Spin an inclined globe with spots rotated in an orbit while not spinning, and both rotated and spun.
The spots form parallel lines perpendicular to the angular velocity vectors.
15.4.2.2 Kick a moving ball
Kick a moving ball to score a goal!
15.4.2.3 High road low road
Race two balls, one down an incline the other down same slight incline that includes a valley.
15.4.2.4 Passing the train
Let a ball accelerate down an incline to pass another ball moving at constant velocity on a horizontal track or pass a striped rope moving at constant velocity in the background and note where the two balls, or the ball and striped rope, have the same instantaneous velocity.
15.4.2.5 Galileo's circle
Simultaneously release small balls to roll down guides that form chords of a large inclined circle to make a single click marking simultaneous arrival.
15.4.2.6 Brachistochrone track
See 2.0.5: Conic sections, parabola.
See 2.0.6: Parabola equation.
Release balls at any height on the brachistochrone track to reach the middle at the same time.
Let balls roll down an incline brachistochrone and a parabola, so that the ball on the brachistochrone wins.
(A brachistochrone track follows a curve that joins two points such that a bead travelling along it under the influence of gravity takes a shorter time than along any other curve.)
Release two balls on opposite sides of a cycloid to always meet in the middle.
The ball on the cycloid always beats the ball on the incline.
(A cycloid is a curve traced by a point on the circumference of a cycle as the circle rolls along a straight line. Oxford English Dictionary.)
Note the use of brachistochrones and cycloids for children in municipal parks.
15.4.2.6.1 Energy required to loop the loop
Demonstrate that a minimum energy is required for a body to complete a vertical loop by releasing a steel ball bearing at different heights on a ramp and note the position where the ball just completes a loop is noted.
Compare the results with 5/2 R.
15.4.2.7 Sliding weights on triangle
Chose different lengths and angles of a wire frame triangle so that two beads sliding down the wires traverse each side in the same time.
15.4.3.1 Lifting
Stick unequal size corks on each end of a knitting needle, place a cord under at the centre of mass and jerk it into the air.
15.4.3.2 Throwing
1. Throw a light disc with attached heavy slug that you can move from the centre to side.
Mark the centre of mass before you throw the disc.
2. Throw a slab of Styrofoam with lights placed at the centre of gravity and away from the centre of gravity.
3. Mark the centre of gravity of a hammer with a white spot.
Throw it in the air.
Attach it to a hand drill to show it rotating smoothly.
4. Tie together a bunch of junk and throw it across the room.
15.4.3.3 Spinning, dry ice puck
Put markers on a large block of wood at and away from the centre of mass.
Place the block on a large sheet of paper and hit off centre with a hammer.
Use a pool cue to hit a dumbbell double dry ice puck on or off the centre of mass.
Use an Earth Moon system hanging from a string to show the earth's wobble.
Rotate the Earth Moon system from a hand drill on and off the centre of gravity.
Attach two unequal masses at the ends of a rigid bar, spin the system about holes drilled in the bar at and off the centre of mass.
15.4.3.4 Centre of mass disc
Use a pool cue to hit a dumbbell double dry ice puck on or off the centre of mass.
15.4.3.5 Air table centre of mass
Stick unequal size knitting needles in a cork. Place, a cord under the cork at the centre of mass and jerk the cork it into the air.
15.4.3.6 Earth moon system
Use an earth moon system hanging from a string is used to show the earth's wobble.
Rotate the earth moon system from a hand drill on and off the centre of gravity.
Attach two unequal masses at the ends of a rigid bar, spin the system about holes drilled in the bar at and off the centre of mass.
15.5.1 Flattening earth
Spin a globe made of flexible brass hoops, or sponge rubber ball, until the hoops flattens and becomes oblate.
Spin a deformable clay or glycerine ball until it flattens or bursts.
15.5.2 Empty jug by swirling Empty a jug when contents are swirled then not swirled to show that the swirled jug empties faster.
15.5.3 Water parabola
See 3.8.0: Conic sections, parabola.
Rotate a rectangular transparent box partially filled with coloured water until you see the parabolic shape.
Spin a glass half filled with coloured water on a rotating table.
15.5.4 Rotate a manometer
Rotate tubing constructed in an E-shape on its back that is partly filled with water.
Note the level of water in the arms.
Rotate a U-shaped manometer with one of its arms coincident with the axis of a rotating table.
15.5.5 Balls in water centrifuge
Spin cork and steel balls in a curved tube filled with water.
Spin a glass bowl containing water, a steel ball and a cork.
Spin a semicircular tube filled with water containing two corks.
Spin one cork is tied to the bottom of a cylinder and one ball is tied to the top of a cylinder full of water at the ends of a rotating bar.
15.5.6 Inertial pressure gradient
Spin a tube containing a bubble in water until the bubble goes to the centre when whirled in a horizontal circle.
15.5.7 Turning motor car, with a candle and balloon
Observe a candle, a carbon dioxide balloon and a hydrogen gas balloon on strings, in a motor car driven in uniform circular motion.
15.5.8 Rotate a doll
Rotate a doll dressed in a full skirt and a doll dressed in a short skirt.
15.5.9 Rotate a candle
Place a lighted candle in a chimney lamp on a rotating table until the flame points to the centre.
Lighted candles in chimneys are rotated about the centre of mass.
15.5.11 Paper saw
Spin typewriter paper at high speed to cut through tougher paper and cardboard.
15.5.12 Rotate a rubber wheel
Rotate a rubber wheel so that the radius stretches.
15.5.13 Rotate a ball on a string
See diagram 15.5.13: Ball on a string.
1. Twirl a ball tried to a string in a vertical direction so students can see the path of the ball.
2. Tie a lightweight ball to a sting and twirl around in a vertical circle.
Use a glass tube for the holder and rubber stoppers for the masses and twirl the masses around your head.
15.6.1 Disc circle with gap, broken ring
Roll a ball around a circular hoop with a gap.
Where will the ball go when it reaches the opening?
15.6.2 Crossing the river
Pull a long sheet of paper along the bench while a toy clock-work car crosses the paper.
The toy car moves across a sheet moving at half the speed of the toy car.
A small stick placed on the top tread of a toy caterpillar tractor moves twice as fast as the toy tractor.
15.6.3 Release ball on a string
Release the string while swinging a ball overhead.
Be careful! Use a slingshot, a David and Goliath type of slingshot
15.6.4 Grinding wheel
Observe direction of sparks flying off a grinding wheel.
15.6.5 Spinning disc
Red drops fly off a spinning leaving traces tangent to the disc.
Place erasers on a disc at various radii and rotate until they fly off.
Falling off the merry-go-round:
A turntable is rotated until objects slide or tip over.
See amusement park horizontal turn table.
Line up coins radially on a rotating platform and spin at varying rates until all fall off.
15.7.01 Projectile motion, projectiles (horizontal and angular projection)
"Da Vinci Catapult" (siege weapon), soft clay balls reach 4 m (toy product)
See diagram 15.7.0: Projectile motion.
See 14.1.0.1: Equations of uniformly accelerated motion, gravity.
1.0 A ball in free fall from height, h
1.1 s = ut + at2/ 2.
1.2 h = 0 + gt2/ 2 = gt2/ 2.
1.3 Time to fall, t = 2h/ g.
2.0 A ball projected from the edge of a table at velocity v
2.2 h = gt2/ 2.
2.3 t = 2h/ g.
2.4 Distance travelled before ball reaches the floor, d, = vt = v (2h/ g).
3.0 A ball projected from the ground with initial v in a direction making angle θ with the horizontal.
3.1.0 Vertical motion
3.1.1 Time of flight = time to reach greatest height × 2
v = u + at, 0 = (v sin θ - gt, ) t = (v sin θ/ g), So, time of flight = 2t = (2v sin θ / g).
3.1.2 Greatest height
(v2 = u2 + 2as), 0 = (v2sin2 θ -2gh), So, greatest height = (v2 sin2 θ/ 2g).
3.2.0 Horizontal motion (horizontal distance, s = range)
3.2.1 Range = (v cos θ × 2t) = (2v2 cos θ sin θ/ g) = (v2 sin 2θ / g).
3.2.2 Maximum range
The sin of an angle is not > 1, so maximum range is possible only when 2 θ = 1, i.e. 2 θ = 90o, and θ = 45o.
So maximum range = (v2/ g), where v = initial velocity of projection.
15.7.1 Projectile cart, projectile launcher, ballistics cart, gun and tunnel, howitzer and tunnel
See: Falling Ballistics, (Commercial).
See diagram 15.7.1.1 Gun and tunnel.
1. Set the trigger on the cart.
Give the cart a strong push towards the tunnel.
If the ball hits the tunnel, you've not pushed the cart hard enough.
Track is 10 feet long.
2. A spring loaded gun in a barrel on a cart shoots a ball vertically and, after the cart passes through a tunnel, the ball lands in the barrel.
3. A ball fired vertically from cart with a barrel mounted on it moving horizontally falls back into the barrel.
The projectile glider consists of a spring loaded vertical cannon enclosed within a funnel.
The cannon is loaded, the firing mechanism is primed and the glider is given a small initial velocity.
A remote control acts as the firing mechanism to propel an object vertically.
The projectile is caught by the funnel.
4. A spring-loaded gun on a cart shoots a ball vertically and, after the cart passes through a tunnel, the ball lands in the barrel.
A ball fired vertically from cart moving horizontally falls back into the barrel.
5. A mobile projectile cart that can project objects vertically is used to show the independence of horizontal and vertical motion.
The projectile cart carries a spring loaded vertical cannon enclosed in a funnel.
A remote control controls a firing mechanism.
The cannon is loaded with a projectile and made ready to fire.
The cart is pushed to give a small initial velocity.
The remote control is used to fire the cannon.
The projectile is caught by the funnel on the cart.
15.7.2 Simultaneous fall
Two balls simultaneously dropped and projected horizontally hit the floor together.
Drop one billiard ball and shoot another out simultaneous.
One ball is projected horizontally as another is dropped simultaneously.
Instructor rolls a ball off the hand while walking at a constant velocity.
15.7.3 Midair target
A hunter shoots a compressed air at a target released when the gun is fired.
The ball hits the target in midair.
15.7.4 Range of a gun
1. Fire a spring loaded gun at various angles.
Use a tennis ball serving machine to find muzzle velocity and range of a gun.
2. Projectile motion: Two angles result in the same launched distance.
If a target on the same plane as the gun can be hit when the barrel of the gun is 30o to the horizontal, the same target can be hit when the barrel of the gun is at 60o to the horizontal (45 -15) = (45 + 15).
The highest point of the flight path of the projectile fired at 60o = 3 × the highest point of the flight path of the projectile fired at 30o.
3. Set the angle to (45 + 20)o and (45 -20)o.
Fire the projectile and observe where it lands.
It lands at the same place.
15.7.5 Parabolic trajectory
See 3.8.0: Conic sections, parabola.
Throw a piece of chalk so it follows a parabolic path drawn on the chalkboard.
Roll ink dipped balls down an incline onto a tilted stage on an overhead projector.
Roll a tennis ball covered with chalk dust across a tilted blackboard.
A stream of water matches the position of balls of lengths 1, 4, 9, 16, at all angles of elevation.
15.8.1 Moving reference frames, relative velocity, independence of horizontal and vertical velocities
15.8.1.1 Crossing the river
Pull a long sheet of paper along the bench while a toy clock-work car crosses the paper.
The toy car moves across a sheet moving at half the speed of the toy car.
A small stick placed on the top tread of a toy caterpillar tractor moves twice as fast as the toy tractor.
15.8.3.5 Coriolis force as an inertia force
Observe the motion in a spinning frame of reference from a static frame of reference.
Coriolis force is a kind of inertia force, i.e. caused by non-inertial system.
For an observer who stand in a frame of reference, spinning at angular speed sa, the object that is moving relative to the frame will be acted by, not only an inertial centrifugal force, but also an inertial force that makes the object not to leave the motion orbit, i.e. Coriolis force.
The direction of the Coriolis force depends on cross product of the speed vector V, relative to the spinning frame and angular speed of the frame itself, always vertical to V and opposite to acceleration of tangent direction.
The magnitude of the Coriolis force is 2mVsa.
Due to the spinning of the earth, the moving object relative to earth is seen acted by a Coriolis force by an observer on earth.
Many phenomena on the ground and in the air in nature are caused by it.
The cold air in high latitude of north, south sphere drops, due to the spinning of the earth, forms a southeast and northeast wind in each sphere.
Due to the air of the equator rises, the two winds can meet near the equator to form a typhoon in the region of south Pacific Ocean.
15.8.3.6 Chalk line on a turntable
The Coriolis force operates independently of the direction in which an object is travelling.
In demonstrations where you draw a line radially across a slowly spinning turntable, it works only when the line is drawn towards or away from the centre.
In the demonstration a chalk line is drawn radially on a slowly spinning turntable.
The track of the chalk is a straight line, but the line drawn on the turntable is curved.
In the inertial frame of the classroom, the track was straight, while in the rotating frame of the turntable the track is curved.
It is easy to draw the line radially, but very hard to match the velocities of the turntable and chalk closely enough by hand to draw a similar line tangentially.
15.8.3.7 Rolling ball on turntable
Roll a ball rolls across a slowly rotating turntable to show the Coriolis effect.
15.8.3.8 Turning sheet
Turn a nearly vertical sheet as a drop of ink is running down it.
15.8.3.9 Walk on turntable
Walk in a straight line on a turntable or a merry-go-round and feel a very strange force.
15.8.3.10 String through spinning globe
Thread a ball on a string through the pole of a spinning globe, then pull on the string and the ball moves to higher latitudes and crosses the latitude lines spinning.
15.8.3.11 Deflected water stream
Mount a can with a hole above a rotating table so that as the table turns the stream of water is deflected.
Stretch a flexible rubber tube with water flowing in across a turntable that can be rotated so that the tube deflects when rotated.