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. 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 zero to the line of action of the force.
Diagram 1.1
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| The two masses are perpendicular to the lever arm. |
Procedure:
1. Balance the meter stick on clamp and record the position of the balance point. What point in the meter stick does this correspond to?
Diagram 1.2
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| the balance point was rounded to 49 cm |
2. Select two different masses 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 caculation? Sum the torques about your pivot point O and compare with the expected value.
Yes the mass of the clamps should be included because they contribute to the total weight of the system.
Diagram 1.3
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| Torque1 - Torque2 = 0 N m |
length 1 = 40 cm = 0.40 m length 2 = 31.16 cm = 0.3116 m
wight 1 = (0.1745 kg)(9.8 m/s^2) weight 2 = (0.2206 kg)(9.8 m/s^2)
= 1.7 N = 2.2 N
Sum of Torques
T1 - T2 = 0 N m
(length 1)(weight 1) - (length 2)(weight 2) = 0 N m
(0.40 m)(1.7 N) - (0.3116 m)(2.2 N) = 0 N m
0.001 N = 0 N
Percent Error:
solve for one of the lengths (L2)
(length 1)(weight 1) - (length 2)(weight 2) = 0 N m
(0.40 m)(1.7 N) - (L2)(2.2 N) = 0 N m
L2 = 0.312 m
((0.3116 m - 0.312 m) / 0.312) * 100 = 0.1 % error
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 the masses and positions. Calculate the net torque on this system about the point support and compare with the expected value.
Diagram 1.4
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| Diagram shows the distance between the center to the weights. The net force of torque equals zero. |
mass1 = 174.5 g = 0.1745 kg mass2 = 270.8 g = 0.2708 kg
length1 = 15.0 cm = 0.15 m length2 = 34.0 cm = 0.34 m
mass3 = 220.6 g = 0.2206 kg
length3 = 30.0 cm = 0.30 m
net torque:
T2 - T1 -T3 = 0
(length 2)(weight 2) - (length 1)(weight 1) - (length 3)(weight 3) = 0 N m
(0.3116 m)(2.2 N) - (0.40 m)(1.7 N) - (0.30 m)(2.16 N) = 0 N m
0.0016 N m = 0 N m
Percent Error:
solve for one length (L2)
T2 - T1 -T3 = 0
(length 2)(weight 2) - (length 1)(weight 1) - (length 3)(weight 3) = 0 N m
(L2)(2.2 N) - (0.40 m)(1.7 N) - (0.30 m)(2.16 N) = 0 N m
L2 = 0.3394 m
((0.3394 - 0.34) / 0.3394) * 100 = 0.2 % error
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.
Diagram 1.4
| ||
| In this step we are solving for the unknown mass of weight 2. |
mass3 = 220.6 g = 0.2206 kg mass1 = 174.5 g = 0.1745 kg mass2 = ???
length3 = 30.0 cm = 0.30 m length1 = 15.0 cm = 0.15 m length2 = 30.2 cm = 0.302 m
solve for the unknown weight (w2)
T2 - T1 -T3 = 0
(length 2)(w2) - (length 1)(weight 1) - (length 3)(weight 3) = 0 N m
(0.302 m)(9.8 m/s^2)w2 - (0.40 m)(1.7 N) - (0.30 m)(2.16 N) = 0 N m
w2 = 0.3043 kg = 304.3 g
percent error:
((304.6 - 304.3) / 304.6) * 100 = 0.01 % error
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?
Diagram 1.5
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| w1 = 199.9 grams at the 90 cm mark |
length2 = 29 cm = 0.29 m
mass of meter stick????
remember that the torques in the system are equal.
T1 = T2
w1*L1 = w2*L2
m1*g*L1 = m2*g*L2
(199.9 kg)(9.8 m/s^2)(0.12 m) = m2(9.8m/s^2)(0.29 m)
m2 = 82.717 g
percent error:
((82.7 g - 81.8 g) / 81.8 g) * 100 = 1.1 % error
6. With the 200 grams still at 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 result.
Diagram 1.6
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| Pictorial representation to solve for x |
m2 = 0.100 kg L2 = 30 cm = 0.30 m
m3 = 0.200 kg L3 = 90 cm = 0.90 m
center of mass = 1/M Σ mi*xi = (m1x1 + m2x2 + m3x3) / (m1 +m2 + m3)
cm = 0.655 m = 65.5 m
percent error:
actual center = 65.1 cm
((65.1 cm - 65.5 cm) / 65.1 cm) * 100 = 0.6% error
Conclusion:
The lab is a helpful way in learning how to calculate torque using the center of balace of the meter stick as an example.There were different cases in calculating the net torque. The first part we just had to balance the meter stick in the center. After we had to place two weights on opposite ends until we found the new balance point. This point was located at 0.4m and 0.3 m from the center. The percent error that we obtained was only a 0.1%. Then we placed a third wight on one end and we did the same. Calculated the torques and set then up to 0 since the system is at equilibrium.
All the percent errors that we calculated were really small meaning that our calculations were accurate. Some sources of error that may occur during the experiment is that the center of balance may be off and it will ruin the data collected.
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