# Homework Help: Rotational acceleration Pully

1. Oct 12, 2007

### royguitarboy

1. The problem statement, all variables and given/known data
Two objects are attached to ropes that are attached to wheels on a common axle as shown below. The two wheels are glued together so that they form a single object. The total moment of inertia of the object is 41 kg·m2. The radii of the wheels are R1 = 1 m and R2 = 0.4 m.

(a) If m1 = 24 kg, find m2 such that there is no angular acceleration of the wheels.

(b) If 12 kg is gently added to the top of m1, find the angular acceleration of the wheels.

2. Relevant equations

T=I(alpha)

alpha=change in velocity/ change in time

3. The attempt at a solution

I'm not really sure where to start for this. I know I need to figure out the Tension for the block, but I'm not sure how to go about doing this.
1. The problem statement, all variables and given/known data

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2. Oct 12, 2007

### Staff: Mentor

Start by labeling all the forces acting on each mass and on the wheel.

Hint for a: It's in equilibrium. What does that tell you about the sum of the forces or the sum of the torques?

Hint for b: Apply Newton's 2nd law to each object and combine the resulting equations.

3. Oct 12, 2007

### royguitarboy

I know that the overall net Force is 0 because there is no acceleration. All I get when I apply Newton's Laws is F=T-mg. This would mean that the Tension equals mg, which I don't think is true. I think I need to figure out the relationship between the two tensions, but I'm not sure how.

4. Oct 12, 2007

### Staff: Mentor

It is most definitely true. (For part a.)
By examining the torques on the wheel. They must total zero as well.

5. Oct 13, 2007

### royguitarboy

Wouldn't the tensions equaling mg mean that m2 would need to equal m1?

6. Oct 13, 2007

### Staff: Mentor

Not at all. Each rope has its own tension: $T_1 = m_1 g$ and $T_2 = m_2 g$. The tension in each rope cannot be equal, since the wheel is in equilibrium. (The ropes attach to the wheel at different radii.)

Last edited: Oct 13, 2007
7. Oct 14, 2007

### royguitarboy

So how do I go about figuring out the tensions?

8. Oct 14, 2007

### Staff: Mentor

For part (a) the tensions are equal to the weights, as I thought you realized.

9. Oct 14, 2007

### royguitarboy

I understand that the tensions equal mass times gravity. I need to figure out what the second mass is, and I'm pretty sure I do this by figuring out how the tensions are related.

10. Oct 14, 2007

### Staff: Mentor

Right. Set the net torque on the wheel equal to zero. That will tell you how the tensions are related.

11. Oct 14, 2007

### royguitarboy

I'm sorry I'm having a hard time understanding this; The tensions must cancel each other since there is no acceleration right? Maybe if it's explained in another way I might get it.

12. Oct 15, 2007

### Staff: Mentor

No. The torques must cancel, otherwise the wheel will accelerate. Torque depends on both force (the tension) and distance from the axis (radius):

$$T_1 R_1 = T_2 R_2$$

13. Oct 15, 2007

### royguitarboy

Ok, I have the first part now. For the second, I get that the T=m1(g+a) and T=m2(g-a), and then that a=(m1g-m2g)/(m1+m2). Is that even close for (b) If 12 kg is gently added to the top of m1, find the angular acceleration of the wheels ?

Last edited: Oct 15, 2007
14. Oct 16, 2007

### Staff: Mentor

No, it's not that simple. The tension in each rope and the acceleration of each mass is different. Remember that you want the angular acceleration of the wheel, not the linear acceleration of the masses. How are they related?