Balancing Torques Lab - Answer to this https://rucsm.org/physics/labdescriptions/1025.pdf PDF

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Answer to this
https://rucsm.org/physics/labdescriptions/1025.pdf...

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Conservation of Momentum - Lab 10 A. Meter Stick Supported at Center of Gravity A. Meter Stick Supported at Center of Gravity Mass of stick: 93.6g Mass of first clamp: Weight=17.5g

Mass of second clamp: Weight=14.0g

Mass of hanger 1: Weight=6.0g

Mass of third clamp: Weight=17.8g Mass of hanger 2: Weight=6.0g

Position of Center of Mass of meter stick: 0.5 m Discussion of how the balance of the meter stick varies as the weights are moved: Even though the meter stick was balanced, the clamps and hangers combined don’t weigh the same so if both moved the same distance away or close to the center of mass then one side of the stick with the heaviest clamp would fall. Therefore to balance the meter stick, the heaviest clamp was moved closer to the center and the third clamp was moved farther from the center. This is because the heaviest clamp has the shortest perpendicular distance from the axis of rotation to line of force, or it’s lever arm, so it’s necessary to move the third clamp to the end of the stick, thus having a higher lever arm because torque is mass times gravity times lever arm. The third clamp would have a higher distance value than the first clamp allowing the third clamp to have a torque value close to the first clamp thus both torques will be about the same and equal to zero. When both torques are zero, the center of mass of the meter stick will stay balanced at .5m. Two Known Weights Data Table (first equilibrium arrangement of weights) Sketch (labeled)

Weights, Position

Lever Arms:

Results:

A

W1= 223.8g X1 = 0.906m W2=523.5g X2 =0.327m

r1=0.406m r2=0.173m

τCCW= 0.89 N τCW= 0.89 N % Diff: 0% How does this compare to the condition required for balance?

Observation:

Instead of focusing on clamps and hangers, now we have weights on the hangers. So we must consider all 3 objects’ weights for counterclockwise and clockwise torque. While weights were 200 and 500 grams respectively, we also had to include the hangers’ and clamps’ weight as well. While the third clamp was heavier, you can see that it was closer to the pivot. This was to lower the lever arm because it already had a high torque value from its weight. The first clamp was farther and had a distance value that made up for it’s low mass in order to have a high torque number that can be equal to the third clamp. We were able to move both clamps in different positions in order to balance the meter stick, meaning we calculated the first clamp’s clockwise torque and the third clamp’s counterclockwise torque that equaled to zero. Three Known Weights Data Table (first equilibrium arrangement of weights) Sketch (labeled)

B

Weights, Position

Lever Arms:

Results:

W1=223.8g X1 = 0.610m W2=123.5g X2 =0.437m W3=78.6g X3 =0.280m

r1=0.110m r2=0.063m r3=0.220m

τ1=0.24 N τ2=0.076 N τ3=0.169 N

τCCW=0.24 N

τCW=0.245 N

% Diff: 2.06%

Additional Discussion: Similar to the second table but now we have a fourth clamp, hanger and weight. As you can see from the table, none of the weights are rounded numbers because each weight included the mass from the slotted mass set, hanger and clamp. Now for each weight there’s a different position because the heaviest weight was set on one side of the stick while the other two were set on another. Since the center is .5m, the closest weight to it was W2, which wasn’t the heaviest because we had to make sure the distance for both weights on one side could amount to a torque value that would be the same from the other. However the clockwise torque had a higher value therefore there was a percent difference. This is probably due to the fact that W3 or W2 could have been positioned differently so that their torque value wasn’t higher than W1’s torque. Or we could have moved W1 differently too so that it could have a greater torque value.

Three Known Weights Data Table (second equilibrium arrangement of weights)

Sketch (labeled)

Weights, Position

Lever Arms:

Results:

C

W1=223.8g X1 = 0.700m W2=123.5g X2 =0.380m W3=78.6g

r1=0.200m r2=0.120m

τ=0.439 N τ=0.145 N

Predicted position and lever arm: X3=0.120m r3=0.382m

Measured position and lever arm: X3=0.107m r3=0.393m

% Diff (lever arms): 2.83%

Discussion and Observations: While this procedure is similar to the last, we now have to predict the position and lever arm of W3. This means that we needed to know W1’s torque and W2’s torque in order to predict what W3’s torque value needed to be and from there consider its mass to find its lever arm. This would make W1r1+W2r2=W3r3, and since every variable, except for r3, is accounted for then we would be able to get .382m for r3. We’d take that value to find the position. Ultimately though the measured position and lever arm wasn’t the same, there was still a percent different which means we must have done our math wrong. Unknown Mass Sketch (labeled)

Weights, Position

Results:

W1=??? X1 =0.380m W2=123.8g X2 =0.760m

W1 (c) =268.23g W1 (m) =245.5g % Diff: 8.84%

Describe calculations needed to determine unknown mass: The unknown mass can be determined by balancing the torques. W2 and X2 are already in place, those were known values that we picked. Therefore we end up with the equation W1r1-W2r2 =0 or W1r1 = W2r2. We also get X1 because we can move both weights at different distances to balance so that their torques equal zero. This means that W1 can be alone on one side, making the equation to be W1 = W2r2/r1. 1. When two unequal masses are balanced on either side of the (balanced) meter stick, what is the relationship between the ratios of the masses and the ratios of their lever arms? When they are balanced then the stick is not rotating. It has no angular acceleration so the net torque is zero. Which means that W1r1 - W2r2 = 0. Then W1/W2 = r2/r1 so the ratio of the masses is the inverse of the ratio of the lever arms. 2. Is it possible that there is no net (total) force on an object (such as the meter stick) but a non-zero torque? Explain and give an example. Yes, it is possible. That means the object itself is not moving, but it is rotating. For example, on a seesaw, one side goes down as the other side goes up and vice versa. The

center of mass of the seesaw does not move at all. However the ends of the seesaw rotate clockwise and counterclockwise. Therefore, there is a net force of zero on the seesaw, but there is non-zero torque when it is rotating. 3. How does the triple-beam balance work? When using a triple beam balance to measure the mass of an object, it uses torque. The balance is occured when the torque due to the measured object equals the total torque applied on the other side by the combination of the three calibrated weights....