Lab #6 - Torques and Center of Mass



Lab #7 - Torques and Center of Mass Name: _______________

Copyright R. Mahoney 2014

This is a relatively straightforward lab designed to show how torques can be balanced, and how the center of mass (CM) of any lever mechanism must be included in any torque calculation involving the lever when the lever's CM is not over the pivot stand (fulcrum).

In step 3, mechanical advantage (MA) for a particular lever system will be calculated.

In step 6, we will show that torques can be calculated relative to any pivot point (or fulcrum), for a system in rotational static equilibrium.

In step 7, we will also show that a stationary lever system must obey the First Condition of Equilibrium too.

Procedure

1. Insert a meter stick into a holding clip somewhat loosely, and mount the stick on the balancing fulcrum. The tightening screw of the clip should be down when mounted in the pivot stand (this is more stable). Also, remove the hanging part of this clip, as it will not be needed. (Call this clip the modified clip.)

Now adjust the meter stick until it balances by itself. The holding clip marker tells you where the natural balancing point (CM) is. What is the reading? Answer: ________ cm

(Don't assume it's the 50 cm mark.)

Now measure the mass of the meter stick, without the modified clip attached.

Answer: ________ gm (Call this mass A.)

Now measure the mass of the modified clip. Answer: ________ gm (Call this mass B.)

2. Measure the mass of a weight holder and an unmodified holding clip plus 100 gm (call this mass 1), and then measure a second weight holder and unmodified holding clip with no added mass (call this mass 2). What are the measured masses?

mass 1: ________ gm mass 2: ________ gm

3. Without disturbing the meter stick from its balanced state, hook mass 1 to the meter stick at the 25 cm mark (location 1), and hook mass 2 to the meter stick at the 90 cm mark (location 2). (Mass 1 should be on your viewing left, and mass 2 on your viewing right.)

Is the new assembly still in balance? Answer: ________

Which side is up (too light)? Answer: ________

Add mass to the light side until balance is reached.

If you added weight to one side, what are the new values for mass 1 and mass 2?

mass 1: ________ gm mass 2: ________ gm

(Use these values for mass 1 and mass 2 in all future calculations.)

Now, with the meter stick balanced, and using the meter stick itself, determine the distances from mass 1 to the CM, and mass 2 to the CM.

Distance, mass1 to CM: ________ cm

Distance,mass 2 to CM: ________ cm

Thinking of the meter stick and weights as a representation of a lever, and considering location 2 as the place where the effort is impressed, calculate the mechanical advantage (MA) for this lever, using

the ratio of the lever arms: ________

the ratio of resistance to effort: ________

(Q1) Are your two calculations of the lever's MA the same? If not, why not?

4. With location1 on your viewing left, and location 2 on your viewing right, calculate the counterclockwise torque (CCWT) due to mass 1, and the clockwise torque (CWT) due to mass 2. Express your answers in gm times cm, instead of changing the gm to newtons, and the cm to m.

CCWT: ________ gm-cm CWT: ________ gm-cm

What is the percent difference between the two torques above? Answer: ________ %

(Q2) Why don't you have to change gm to newtons, and the cm to m?

(Q3) Why doesn't the mass of the meter stick and its modified clip matter in the above torque calculation?

(Q4) Why are we not concerned about the (sometimes slightly non-perpendicular) angles the weight holders make with respect to the meter stick?

5. Start over, but this time move the modified clip on the meter stick to the 30 cm mark on the meter stick. Now place the meter stick's modified clip on the pivot stand. (The long side of the meter stick should be to your viewing right.)

Now add a holding clip plus weight holder on the 10 cm side of the stick.

Does it balance? Answer: ________

If not, add mass to the weight holder at the 10 cm mark until the meter stick balances.

Once balance is achieved, remove the holding clip, weight holder and added weight (if any), and weigh them all together (call this mass 3).

What is the weight of mass 3? Answer: ________ gm

What is the distance from the new pivot point, the 30 cm mark, to the CM?

Answer: ________ cm (Should be around 20 cm.)

What is the distance from the new pivot point to mass 3?

Answer: ________ cm (Should be around 20 cm. Call this distance D.)

With mass 3 on your viewing left, calculate the counterclockwise torque (CCWT) due to mass 3. With the CM on your viewing right, calculate the clockwise torque (CWT) due to the meter stick itself (think of the stick's mass as being at the CM). Express your answers in gm times cm.

CCWT: __________ gm-cm CWT: __________ gm-cm

What is the percent difference between the two torques above? Answer: ________ %

(Q5) How does the weight of the meter stick (mass A) compare to the weight added at the 10 cm mark (mass 3)?

6. Don't break down your setup from step 6.

Consider the lever system used in step 6. Since it is in rotational static equilibrium, you can in theory place an imaginary fulcrum anywhere under the lever, and it will be in balance relative to that imaginary fulcrum. So now, in your mind, place an imaginary fulcrum under the 10 cm location, and with the 10 cm mark on your viewing left, see the meter stick mass (mass A) applying a clockwise torque (CWT), and the actual pivot stand applying a counterclockwise torque (CCWT) relative to the imaginary fulcrum.

What is the distance from the 10 cm mark to CM? Answer: ________ cm

(Should be about 40 cm. Call this distance L.)

Calculate the meter stick's CWT (mass A times distance L). Answer: ________ gm-cm

(Call this Torque A.)

What is the distance from the 10 cm mark to the actual pivot stand?

Answer: ________ cm (Should be around 20 cm.)

Using the Second Condition of Static Equilibrium (balanced state of CWTs and CCWTs), determine what upward force the actual pivot stand must be impressing on the meter stick at the 30 cm mark. While is is understood that the answer to this question should be in newtons, express your answer in gm, since Torque A is in gm-cm.

Answer: ________ gm (Call this mass C.)

(Q6) How does mass C compare to mass A?

7. From the First Condition of Static Equilibrium, the upward force the actual pivot stand impresses on the meter stick should equal the downward weight of the entire lever system. Put another way, mass C should equal mass A plus mass 3. Is this true?

9. State two non-trivial systematic errors for this experiment.

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