Download Torque and Rotational Equilibrium

Torque and Rotational Equilibrium
Name
Section
Theory
Torque is the rotational analog of force. If you want something to move (translate), you apply a force; if you
want something to rotate, you apply a torque. Torque is defined as
τ = Fd
(1)
where F is a force and d a distance. This distance is known as the moment (or lever) arm of the torque
and is defined as the perpendicular distance from the line of action of the force to the axis of rotation. The
SI unit of torque is a Nm. By convention, torques which would produce a clockwise rotation are considered
negative; torques which would produce a counterclockwise rotation are considered positive.
If a system is in equilibrium, then two conditions must be met:
1. The sum of the forces (the net force) on the system is zero.
2. The sum of the torques (the net torque) about any point is zero.
Figure 1: A System in Equilibrium
Consider the system shown in Figure 1. A meter stick is supported at its center of gravity (CG) with a mass
suspended from each end. The first condition states that the sum of the forces on the system is zero. Here,
the sum of the weights of m1 and m2 (both acting downward) is equal to the force of the support on the
meter stick (acting upward); i.e.,
Fsupport − m1 g − m2 g = 0.
The second condition states that the sum of the torques about any point must be zero. If we sum torques
about the support point, then the counterclockwise torque due to m1 is equal to the clockwise torque due
1
to m2 ; i.e.,
F1 d1 − F2 d2
=
0
m1 gd1 − m2 gd2
=
0.
The upward force on the meter stick at the support contributes no torque about this point because d = 0!
Figure 2: The Same System
However, notice that this condition states that the sum of the torques about any point is zero. If we instead
sum torques about the left end of the meter stick (Figure 2), then
−m1 gd1 + Fsupport dsupport − m2 gd2 = 0.
The force of the support on the meter stick will contribute a torque about the left end of the meter stick
(d 6= 0). Look at it this way - if Fsupport were the only force acting on the meter stick, it would rotate the
meter stick counterclockwise about the left end.
Keep this in mind. The moment arm for a torque changes when you change the point about which you are
summing torques (the axis of rotation). Also, it is possible that you will have a different number of torques
about different points in a system. Finally, a torque that would produce a clockwise rotation about one
point may produce a counterclockwise rotation about another point.
You may be wondering why the weight of the meter stick was not used in any of the calculations above.
Certainly it is another force acting downward, and it would certainly contribute a torque about the left
end of the meter stick. While both are true, it does not matter. The force is trivial; the total force acting
downward would increase, but the upward force of the support on the meter stick would increase by the
same amount; i.e., the sum would still be zero. As for torque, we have an analogous situation because the
meter stick is supported at its CG. The clockwise torque about the left end of the meter stick due to the
weight of the meter stick itself would be exactly offset by the larger counterclockwise torque about this point
due to the increase in the upward force of the support on the meter stick.
Apparatus
Meter stick, Knife-edge clamp, Support, Hooked Masses, String, Scissors, Triple-beam balance.
Procedure
The system you assemble in each of the following procedures starts with the meter stick balanced on the
support by itself. The hooked masses are suspended from the meter stick with loops of string.
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Two Masses in Equilibrium
1. Suspend a 100g mass from the meter stick on one side of the support and a 200g mass on the other side.
Adjust the locations of these masses until equilibrium is achieved (the meter stick is again balanced).
Place the data in Table 1, then calculate the torques about the support point.
Location of support (cm)
Location of 100g mass (cm)
Location of 200g mass (cm)
Force
(N)
Moment Arm
(m)
Torque
(Nm)
100g mass
200g mass
Table 1: Two Mass System; Torques about Support
Three Masses in Equilibrium
1. Suspend a 100g mass and a 50g mass on one side of the support (different locations), and a 200g mass
on the other side. Adjust the locations until the system is again balanced. Place the data in Table 2,
then calculate the torques about the support point.
2. In Table 3, calculate the torques about the left end (0cm) of the meter stick.
3. In Table 4, calculate the torques about a third point of your choosing in the system (other than the
previous two). Note: if one of your moment arms (hence torque) is 0, record it as such.
Location of support (cm)
Location of 50g mass (cm)
Location of 100g mass (cm)
Location of 200g mass (cm)
3
Force
(N)
Moment Arm
(m)
Torque
(Nm)
50g mass
100g mass
200g mass
Table 2: Three Mass System; Torques about Support
Force
(N)
Moment Arm
(m)
Torque
(Nm)
50g mass
100g mass
Support
200g mass
Table 3: Three Mass System; Torques about Left End of Meterstick
Location of point about which you are summing torques (cm)
Force
(N)
Moment Arm
(m)
Torque
(Nm)
50g mass
100g mass
Support
200g mass
Table 4: Three Mass System; Torques about a Third Point
Determining an Unknown Mass
1. Suspend a 200g mass on one side of the support and the unknown mass on the other side. Adjust
the locations until the system is again balanced. Place the data in Table 5. Using that fact that the
system is in equilibrium, calculate the mass of the unknown.
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2. Determine the mass of the unknown with the beam balance for comparative purposes.
Location of support (cm)
Location of 200g mass (cm)
Location of unknown mass (cm)
Force
(N)
Moment Arm
(m)
Torque
(Nm)
200g mass
Unknown mass
Mass of unknown from torques (kg)
Mass of unknown from triple-beam (kg)
Percent error in masses
Table 5: Determination of Unknown Mass
Analysis
1. For the two masses in equilibrium procedure, what was the net torque about the support point? What
should be the net torque about this point?
2. For the three masses in equilibrium, what was the net torque about the support point? What should
the net torque be about this point?
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3. How did the net torque about the other two points you determined compare with the net torque about
the support point in the three mass system? Is this surprising? Explain.
4. How close were you in determining the mass of the unknown from the summation of torques? Are you
confident that a mass could be determined via this method?
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Pre-Lab: Torque and Rotational Equilibrium
Section
Name
Answer the questions at the bottom of this sheet, below the line - continue on the back if you need more
room. Any calculations should be shown in full.
1. What is the clockwise torque about the support point in the system shown above?
2. What is the counterclockwise torque about the support point in the system shown above?
3. Is the system shown above in equilibrium? How do you know?
4. How many torques are there about the left end of the meter stick (0cm) in the system shown above?
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