Torque and Rotational Equilibrium 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 = F d (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 m 1 and m 2 (both acting downward) is equal to the force of the support on the meter stick (acting upward); i.e., F support m 1 g m 2 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 m 1 is equal to the clockwise torque due to m 2 ; i.e., 1
F 1 d 1 F 2 d 2 = 0 m 1 gd 1 m 2 gd 2 = 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 m 1 gd 1 + F support d support m 2 gd 2 = 0. The force of the support on the meter stick will contribute a torque about the left end of the meter stick (d 0). Look at it this way - if F support 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-edged clamp, Support, Hooked Masses, String, Scissors, Triple-beam balance. 2
Procedure Two Masses in Equilibrium 1. Attach the knife-edged clamp to the meter stick and place the assembly on the support, knife-edges against the support. Adjust the location of the clamp until the meter stick balances on its own. Record this support point location. 2. Suspend a 100g mass from the meter stick on one side of the support and a 200g mass on the other side using loops of string. Adjust the locations of these masses until equilibrium is again achieved. Record the location of each mass on the meter stick; i.e., the point is was actually suspended (0cm-100cm). Three Masses in Equilibrium 1. Suspend a 100g mass and a 50g mass on one side of the support at different locations, and a 200g mass on the other side. Adjust their locations until the system is again balanced. Record the locations of the masses (0cm-100cm). Determining an Unknown Mass 1. Suspend a 200g mass on one side of the support and the unknown mass on the other side. Adjust their locations until the system is again balanced. Record the locations of the masses (0cm-100cm). 2. Determine the mass of the unknown with the beam balance for comparative purposes. Mass of the Meter Stick In this procedure, the meter stick is not supported at its CG; therefore, it will contribute its own torque. 1. Move the location of the knife-edged clamp to somewhere between 25cm-40cm on the meter stick. Record this location. 2. Use a single hooked mass to again achieve equilibrium. Record this mass and its location (0cm-100cm). 3. Determine the mass of the meter stick by itself with the beam balance for comparative purposes. Analysis Since the masses are in g and the distances are in cm, use cgs units instead of mks. The force for each torque is the weight (w = mg) of the mass. The mks unit of force is the N kg m/s 2 ; in cgs, it is the dyne g cm/s 2. Torque is τ = F d, so the mks unit is a N m; in cgs, it is a dyne cm. Two Masses in Equilibrium 1. Calculate the torque about the support point for each of the forces. Remember that you have the absolute locations (0cm-100cm) of each mass, but the moment arm for each torque is the distance from the force to the axis of rotation. 2. Calculate the net torque (sum of these torques) about the support point. 3. What should the net torque be about the support point? Are you close enough? Support your answer quantitatively. 3
Three Masses in Equilibrium 1. Calculate the torque about the support point for each of the forces. 2. Calculate the net torque about the support point. 3. Calculate the torque about the left end of the meter stick (0cm) for each of the forces. 4. Calculate the net torque about the left end of the meter stick. 5. Calculate the torque about a third point of your choosing (other than the two already tested) for each of the forces. 6. Calculate the net torque about your third point. 7. How do the net torques about the three points compare? Is this surprising? Explain. Determining an Unknown Mass 1. Using the fact that the system is in equilibrium, calculate the mass of the unknown. Although in theory it would not matter about which point in the system you summed torques, the calculation is easiest about the support point; why? 2. How close were you in determining the mass of the unknown from the summation of torques? Are you confident that a mass can be determined via this method? Mass of the Meter Stick 1. Assume that the entire weight of the meter stick is acting downward at its CG. Using the fact that the system is in equilibrium, calculate the mass of the meter stick. 2. How close were you in determining the mass of the meter stick from the summation of torques? Is it surprising that we could assume that the entire weight of the meter stick acted at its CG? 4
Pre-Lab: Torque and Rotational Equilibrium Name Section 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? 5