EXPERIMENT M5
CENTRIPETAL FORCE
Objectives:
•
•
•
•
To improve your understanding of circular motion
To verify the expression for the centripetal force, F = mv2/r
To determine the relationship between the period of rotation and the centripetal force
To determine the relationship between the period of rotation and the orbit radius
Background:
Objects moving at constant speed in a circular path are one of the first excursions scientists
make towards the general case of motion along a curved path. It is immediately clear that
motion in curved paths is extremely important in most aspects of dynamics. Nearly all objects
will follow a curved path of some sort, e.g. cars going round a bend, golf balls in flight etc.
While the case of an object moving with constant speed around a perfect circle may seem like a
pretty specialised case, in fact it serves well to demonstrate important aspects of motion in a
curved path. We can gain a sound qualitative understanding of this situation simply by
appealing to Newton’s laws. If we look at an object moving in a circular path at constant
speed (which means it goes around equal segments of the circle in equal times no matter where
it is on the circle) at two different times as shown in Figure 1 below:
V1
V2
Figure 1. Two objects in circular motion.
What we can clearly see is that the two velocity vectors ,V1 and V2, are different! They point
in different directions even though they have the same length (i.e. the car has the same speed.
This means that the car’s velocity is continually changing, i.e. the car is accelerating.
We know from Newton’s second law that a force is required to change an object’s velocity, F
= ma, and hence we can see that a force must be exerted on the car to cause it to move in a
circular path. So even though the car’s speed is unchanged, a force is acting on it to change its
velocity vector direction. This force is known as the CENTRIPETAL FORCE. We experience
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this force in everyday life when we travel in a car for example. When the car rounds a bend, we
experience a sensation of being thrown outwards towards the side of the car. In fact, this is an
illusion, because what is happening is that the car is accelerating and pushing inwards on us.
What is the origin of this centripetal force? That depends on the particular case under study.
In the case of the car above, it is the friction between the tyres and the road. In the case of the
earth going round the sun it is the gravitational pull between the earth and the sun which
provides the centripetal force. If you were to attach a weight to a rope and swing it in a circle
above your head, the tension in the rope provide the centripetal force. No matter what its
origin however, anytime an object moves in a curved path it must be experiencing a centripetal
force of some kind. In general, the exact magnitude and time dependence of the centripetal
force are hard to determine, but in the simplest case we will examine, that of an object moving
at constant speed in a circular path, we can determine the magnitude and direction of the force
quite easily. We will find that it always point towards the centre of the circle about which the
object is moving, hence the title centripetal. We will examine how the centripetal force
depends on the circle radius and the object’s speed.
Experimental set-up:
You will be using the centripetal force apparatus shown in the photograph below:
Apparatus
-Turntable with centripetal
force frame attached,
-Air bearing control unit
-Two sliding mass satellites
-One collar
-Air blower
-10 gram mass hanger
-Four 10 gram masses
-Stop clock
This apparatus consists of a turntable with an air supply unit. The air supply unit, the large
green cylinder on the floor, is connected into the side of the turntable and blows air into the
system. This allows the top plate of the turntable to float on a cushion of air (like an air
hockey table) and hence the top table can rotate freely with almost no friction. Sliding masses
(called satellites) may move along a frame connected to the turntable top plate, and weights
can be attached to a hanger in the centre of the frame, and can exert a force on the masses. With
this apparatus the applied centripetal force may be easily varied and the corresponding values
of orbital radius and period of rotation measured. The rotating mass may be regarded as a
‘satellite’ in orbit due to the centripetal force supplied by the tension in the string holding the
mass and a smaller contribution due to friction between the mass and the rail.
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Now answer questions A1 and A2 on the answer sheet
Investigation of the relationship between centripetal force and period of rotation
Procedure:
1. Preparation:
• Unscrew the top rod from the centripetal force frame. Slip the larger sliding mass
"satellite" and a collar onto the top rod, then fix the rod firmly back onto the frame.
• On the top crossbar of the force frame the distance from the centre or axis of rotation is
marked in centimetres. Slide the mass satellite to the position furthest from the axis of
rotation (either 18 or 20 cm, but measure this carefully yourself, part A3 of answer sheet).
Screw the collar to the rod to stop the mass satellite moving towards the axis of rotation.
• Stretch a rubber band between the vertical sides of the frame. Tie a length of thread from
the satellite over the pulley wheel to the mass hanger so that the hanger is on the axis of
rotation. Adjust the rubber band so that it is at the same level as the mass hanger.
2. Starting up:
• Turn on the air blower.
• The air bearing control unit is the variable supply voltage to the motor of the turntable.
Switch it on and slowly increase the supply voltage until the disc is rotating at such a
velocity that the satellite just begins to slide outwards. Observing the gap between the
mass hanger and the rubber band may help to determine when the satellite just begins to
slide.
3. Making measurements:
• Measure the time for 20 revolutions of the disc and then stop the disc. Calculate the
period (T). Record the weight of the mass hanger (weight, W = mass x acceleration due to
gravity, g = 9.81ms-2) and the period of rotation (T).
• Repeat the experiment keeping the radius of the orbit constant but increase the hanging
mass in 10 g stages.
4. Data presentation:
• Carefully tabulate your results (part A4 of answer sheet), with the weight of the mass
hanger, W, in Newtons and the period, T, in seconds.
• Plot the graph of W versus 1/T2 (part A5 of answer sheet) and determine the slope of this
graph.
• Using equation (3) in the analysis
section below we see that this graph should be a straight
4π 2 r.m
line with a slope equal to
. Since we know the mass, m, and the radius, r, we may
then verify the expression for the centripetal force by comparing the value calculated with
equation (3) with the experimental value of the slope (parts A6 and A7 of answer sheet).
• The intercept gives the frictional resistance on the satellite, FFR (part A8 of answer sheet).
• Estimate sources of errors in A9 on answer sheet
Appendix 1 describes how to plot graphs and to estimate errors.
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Analysis:
Newton's first law states that a body will either remain at rest or in a state of uniform linear
motion unless a force acts upon it. Thus, when a body moves in a circular orbit an unbalanced
force must act since the direction of motion of the body and therefore its vector momentum
are continuously changing. It can be shown that the magnitude of this force is given by:
F = mv2/r
(1)
where m is the mass of the rotating body, v its linear speed and r the orbital radius. This force
is directed towards the centre of the circle, hence the term centripetal (centre seeking) force.
Since v = 2π r/T, where T is the period, Eqn. (1) may be rewritten as:
F = 4π2rm/T2
(2)
From this expression it follows that as F increases the period should rapidly decrease and a
graph of F versus 1/T2 should be a straight line.
The centripetal force is supplied both by the weight of the hanging mass, W, and the frictional
resistance on the satellite. Thus from Eqn. (2), W + FFr = 4π2r.m/T2
W = 4π2rm/T2 - FFR
(3)
Equation. (3) is of the form y = mx + c, the equation of a straight line if we choose y = W and
x = 1/T2.
W = 4π 2 .r.m.(
y
= m .
1
T
2
) - FFR
x + c
So a graph of W versus 1/T2 should be linear and have a slope 4π2rm and an intercept on the
y-axis equal to –FFR.
Investigation of the relationship between centripetal force and orbital radius
Procedure:
1. Preparation:
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• Replace the satellite used in part 1 with the smaller one. Keep the hanging mass constant
at 30 g i.e. 10 g mass hanger plus two 10 g masses.
2. Starting up:
• Put the mass satellite at the 10 cm position and affix the collar. As in part 1, determine the
rotation rate required to make the satellite just begin to move outwards. It may be
necessary to shorten the length of thread to prevent the hanging masses beginning to swing
themselves.
3. Making measurements:
• Record the orbit radius, r. Measure the time for 20 revolutions of the disc and then stop the
disc. Calculate the period (T).
• Repeat this for satellite positions in the 10-20 cm range. Again, you may need to shorten
the length of thread particularly at larger radius values.
4. Data presentation:
• Carefully tabulate your results (part A10 of answer sheet), with the orbit radius, r, in
metres and the period, T, in seconds.
• Plot the graph of r versus T2 (part A11 of answer sheet) and determine the slope of this
graph.
• Using equation (4) in the analysis section below we see that this graph should be a straight
line through the origin with a slope equal to
(W + FFR )
4π 2 m
. Since we know the mass, m, and
hence the weight W, we may then find the value of the friction FFR (part A12 of answer
sheet). This should have the same value as was found in the previous experiment.
Appendix 1 describes how to plot graphs and to estimate errors.
Analysis:
From equation. (3) of the analysis section of the previous experiment we have, writing the
centripetal force as W + FFr
r=
y
( W + FFR )
2
4π m
T2
(4)
= m . x
Equation. (4) is of the form y = mx , the equation of a straight line through the origin if we
choose y = r and x = T2.
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So a graph of r versus T2 should be linear with a slope
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(W + FFR )
4π 2 m
and go through the origin.