PHYSICS LABORATORY
Western Kentucky University
Department of Physics & Astronomy
ROTATIONAL EQUILIBRIUM
Purpose
The purpose of this experiment is to study the conditions for rotational equilibrium of a body and to
use those conditions to find an unknown weight.
Theory
Torque
If the vector sum of all the forces acting on a body is zero, the body is in translational equilibrium.
However, the body will not necessarily be in rotational equilibrium. Rotational equilibrium
depends not only on the magnitude and direction of each force, but also on its distance from the axis
of rotation.
Fig. 1 shows a rod which can rotate around axis O.
(In your experiment the rod will be a meter stick and
it will rotate in a vertical plane.) At Point A, force F
is applied, at an angle θ to the rod. In Fig. 1, d
represents the perpendicular distance from axis O to
the line of action of force F; d is called the "moment
arm" or the "lever arm" of force F around axis O.
The tendency of a force to produce rotation is called torque and is represented by the Greek letter τ
(tau). The torque of a force around an axis is the product of the force multiplied by the moment arm.
In Fig. 1 the torque of force F around axis O is τ = Fd = F (OA) sin θ. (Where OA represents the
length of the line segment from point O to point A.)
In Fig. 1, force F tends to rotate the rod around O in a
counter clockwise sense. Fig. 2 shows a different
force F' acting on the rod at point B. F' tends to rotate
the rod in the clockwise sense. The torque of force F'
is τ' = F'd' = F' (OB) sin θ'.
Suppose both F and F' act on the rod. If we call τ
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positive and τ' negative, then the algebraic sum of τ and τ' is the net torque acting on the rod.
*If the net torque acting on a body is zero, then the body is in rotational equilibrium, with no
tendency to rotate in either sense.
Center of Gravity
In Fig. 3, even without force F and F', the rod will tend to rotate around axis O because of its own
weight W. Every piece of the rod contributes to the
weight, of course. But as far as rotation is concerned,
we can treat the weight W as if it acts at one point,
represented by C in Fig. 3. That point is called the
center of gravity of the rod.
Procedure
Before performing the main experiment,
you will need to find the weight W of the rod and the
location of its center of gravity. To find the weight
of the rod is easy enough: weigh it with a balance.
But remember that weight is a force measured in
newtons (N).
to find the location of the center of gravity of the rod, lay it on a knife edge and adjust its position
until it balances, as in Fig. 4. The point where the rod balances is its center of gravity. You
probably will find that the center of gravity of the rod is not far from its mid-point.
Main Experiment
You will use the condition for rotational
equilibrium to find the weight L of a load
suspended as in Fig. 5 from a rod.
1.
The rod (a meter stick) has holes
drilled through it at 10 cm intervals.
A horizontal axis-rod is attached to the vertical support at O. Slip the "10 centimeter hole" onto the axis-ro
2.
Use a spring-balance to exert a force
F on the meter stick at the "70 cm
hole," marked A in Fig. 5. Arrange
the balance so that the angle θ between force F and the meter stick is somewhere between
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60E and 80E.
3.
A small bucket is provided. Hang the bucket on the meter stick at the "90 cm hole," marked
B in Fig. 5.
4.
Slowly add small metal pellets to the bucket while keeping an eye on the spring-balance.
Continue adding metal pellets to the bucket until the reading of the spring-balance is
somewhere between 10 N and 20 N.
5.
Measure the angle θ between the meter stick and the direction of force F. Measure also the
angle θ' between the meter stick and the vertical direction.
6.
You have already determined the weight W of the meter stick and the location of its center of
gravity C.
Data Analysis
1.
Calculate the magnitude of the torque τF of force F around axis O. Express the result in
meter-newtons (mAN).
Is τF clockwise or counter-clockwise? Therefore shall we call it + or - ?
2.
Calculate the magnitude in mAN of the torque τW of force W around axis O.
Is τW clockwise or counter-clockwise? Therefore shall we call it + or - ?
3.
Let L represent the (unknown) weight of the loaded bucket and write an expression for the
torque τL of force L around axis O. Is it + or - ?
4.
Since the meter stick is in rotational equilibrium, the algebraic sum of τF, τW, and τL must be
zero. In other words, TF + Tw + TL = 0, or F d+ W d+ Ld = 0. Recalling that Weight = mg
you should be able to calculate the weight of the loaded bucket (L).
Use this fact to calculate the weight L of the loaded bucket.
5.
Make a direct measurement of weight L by means of a balance.
Results
Calculate the percent difference between your measured value for L and the calculated value for L.
Do your results convince you that the stated condition for rotational equilibrium is correct?
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