Balanced Torques and Center of Gravity

Purpose: To investigate the conditions for rotational equilibrium of a rigid bar and to determine the center of gravity of a system of masses.

Equipment: Meter stick, meter stick clamps (knife edge clamp), balance support, mass set, weight hangers, unknown masses, balance.

Introduction: The condition for rotational equilibrium is that the net torque on an object about some point in the body, O, is zero. Remember that the torque is defined as the force times the lever arm of the force with respect to the chosen point O. The lever arm is the perpendicular distance from O to the line of action of the force.

Procedure:

Note: In each of the following steps, where appropriate, make a careful sketch showing the meter stick with the applied forces and mark their locations. Also, show the point, O, about which you are calculating torques.

1.      Balance the meter stick in the knife edge clamp and record the position of the balance point. What point in the meter stick does this correspond to?

 

2.      Select two different masses (100 grams or more each) and using the meter stick clamps and weight hangers, suspend one on each side of the meter stick support at different distances from the support. Adjust the positions so the system is balanced. Record the masses and positions. Is it necessary to include the mass of the clamps in your calculations? EXPLAIN! Sum the torques about your pivot point O and compare with the expected value.

Mass1= 150g   position1=58.5cm

                                                                                        Mass2= 50g     position2=24.1cm

                                                                                        Mass of clamp:20g

We should include the mass of the clamps.  Because it’s a hole system without it would be missing a

piece of the system.   torque_left =  W₁ × L_l = [(50+ 20)/1000]kg × 9.8m/s^2 × [(48.5- 24.1)/100]m = 0.167 N·m
      torque_right = W₂  × L₂ = [(150+ 20)/1000]kg × 9.8m/s^2 × [(58.5- 48.5)/100]m = 0.166 N·m
      difference of torque = 0.167-0.166= 0.001 N·m
   
      expected value of L₂:
      W₁·L_l = W₂·L₂ 
     [(50+ 20)/1000]kg × 9.8m/s^2 × [(48.5- 24.1)/100]m=  [(150+ 20)/1000]kg × 9.8m/s^2 × [L₂/100]m
      L₂= 10.047cm
      experimental value= 58.5cm-48.5cm= 10cm     
      percent of error= [(10.047-10)/10.047]× 100%= 0.468%

3. Place the same two masses used above at different locations on the same side of the support and balance the system with a third mass on the opposite side. Record all three masses and positions. Calculate the net torque on this system about the point support and compare with the expected value.  mass₁: 150g   position₁: 68.5cm
      mass₂: 50g   position₂: 58.5cm
      mass₃:  100g   position₂: 15.3cm
      mass of clamp: 20g

   
      torque_right = W₁·L_l+ W₂·L₂
      [(50+ 20)/1000]kg × 9.8m/s^2 × [(58.5- 48.5)/100]m+  [(150+ 20)/1000]kg × 9.8m/s^2 × [(68.5-48.5)/100]m = 0.4018 N·m

       torque_left = W₃·L₃
       [(100+ 20)/1000]kg × 9.8m/s^2 × [(48.5-15.3)/100]m = 0.3904 N·m
   
       difference of torque = 0.4018-0.3904= 0.0114 N·m
     
       expected value of L₃:
       W₃·L₃= W₁·L_l+ W₂·L₂
       [(100+ 20)/1000]kg × 9.8m/s^2 × L₃=  0.4018 N·m
       L₃= 34.17cm

       experimental value of L₃= 48.5-15.3= 33.2cm

      
       percent of difference= [(34.17-33.2)/34.17] × 100%= 2.84%

4.Replace one of the above masses with an unknown mass. Readjust the positions of the masses until equilibrium is achieved, recording all values. Using the equilibrium condition for rotational motion, calculate the unknown mass. Measure the mass of the unknown on a balance and compare the two masses by finding the percent difference.

mass₂: 50g   position₂: 58.5cm
      mass₃:  unknown   position₂: 4.8cm
      mass of clamp: 20g

      expected value of mass₃(unknown):
      W₁·L_l+ W₂·L₂ = W₃·L₃
      [(50+ 20)/1000]kg × 9.8m/s^2 × [(58.5- 48.5)/100]m+  [(150+ 20)/1000]kg × 9.8m/s^2 × [(68.5-48.5)/100]m = [(m₃+ 20)/1000]kg× 9.8m/s^2 × [(48.5-4.8)/100]m 
      m₃= 94g
   
      measured value of the unknown mass: 91.5g
   
      percent of error= [(94-91.5)/94] × 100%= 2.66%

5. Place about 200 grams at 90 cm on the meter stick and balance the system by changing the balance point of the meter stick. From this information, calculate the mass of the meter stick. Compare this with the meter stick mass obtained from the balance. Should the clamp holding the meter stick be included as part of the mass of the meter stick? EXPLAIN! balance point: 77.8cm

the mass of meter stick obtained from the balance: 84.2g

the calculated value of meter stick:
W₁·L_l  =  W₂·L₂
m × 9.8m/s^2 × [(77.8- 48.5)/100]m = 0.2kg × [(90- 48.5)/100]m
m= 83.28g

percent of error: [(84.2-83.28)/84.2] × 100%= 1.09%

Because  should still use the 48.5cm  as the balance point the mass of the clamp should still be included.

6. With the 200 grams still at the 90 cm mark, imagine that you now position an additional 100 grams mass at the 30 cm mark on the meter stick. Calculate the position of the center of gravity of this combination (two masses and meter stick). Where should the point of support on the meter stick be to balance this system? Check your result by actually placing the 100 g at the 30 cm mark and balancing this system. Compare the calculated and experimental results.

 

Let the unknown point equal to x cm.

experimental value: x=65.1cm

calculated value:
W₁·L_l+ W₂·L₂ = W₃·L_{3
[100/1000]kg × 9.8m/s^2 × [(x- 30)/100]m+  [(84.2)/1000]kg × 9.8m/s^2 × [(x-48.5)/100]m}
 = [200/1000]kg× 9.8m/s^2 × [(90-x)/100]m 

x=65.29cm 

percent of error: [(65.29-65.1)/65.1] × 100%= 0.29%

Conclusion:  we learned how to find the center of gravity of the system. And how to find rotation of equilibrium. I also learned first to balance to find the center of mass. Then weigh  the entire thing and therefore find the center of gravity. I also learned that  the magnitude of torque depends on the force applied ,the length of the arm and the angle between the force vector and the arm.

Source of error: mass was no accurate because of the way the we weight the ruler. Also the stick is not exactly horizontal therefore it is not 90 degrees.