# Homework Help: Spool of thread physics question

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1. Mar 19, 2016

### frillybob

1. The problem statement, all variables and given/known data
A spool of thread consists of a cylinder of radius R1 = 6.6 cm with end caps of radius R2 = 9.3 cm as depicted in the end view shown in the figure below. The mass of the spool, including the thread, is m = 290 g. The spool is placed on a rough, horizontal surface so that it rolls without slipping when a force = 0.690 N acting to the right is applied to the free end of the thread. For the moment of inertia treat the spool as being a solid cylinder of radius R1, as the extended edges are thin and therefore light.

(a) What is the acceleration of the spool? Take positive to be to the right.

http://www.webassign.net/serpse8/10-p-085.gif

2. Relevant equations
Solid Cylinder I = 1/2(M)(r)^2

3. The attempt at a solution

I need some help with this problem. I don't think my attempt is even correct. But my thought is since T = I*A. Then I just need to solve for A. A = T*I. The moment of inertia is what is stumping me. I'm not sure if I'm doing this correctly. I = 1/2(.3)(.066) = 0.0099. This does not solve the problem adequately. Thanks of the help!

*Where T is force
A is angular acceleration
I is moment of inertia.

Last edited: Mar 19, 2016
2. Mar 19, 2016

### ehild

What is the question?

3. Mar 19, 2016

### frillybob

Updated in main post. Sorry about that!

4. Mar 19, 2016

### ehild

What are A and a? T is the force in the problem. Better to denote it F.

5. Mar 19, 2016

### frillybob

Updated in main post. I left it as T because that is what the problem defines it by.

6. Mar 19, 2016

### rcgldr

The horizontal surface applies a force to the left at the edges of the end caps as the spool accelerates. This torque is partially opposed by the thread's force to the right at the edge of the inner part of the spool.

7. Mar 19, 2016

### ehild

The torque is defined with respect a point or axis. What is your point of reference?

8. Mar 19, 2016

### frillybob

The question says to take positive to be the right. It is moving to the right as well.

9. Mar 19, 2016

### ehild

How do you calculate torque?

10. Mar 19, 2016

### frillybob

T=I*A

Where T is torque (Given)
I is moment of inertia (Not given)
A is angular Acceleration (Need to find)

11. Mar 19, 2016

### ehild

Yes, but how is the torque of a force defined?

12. Mar 19, 2016

### frillybob

13. Mar 19, 2016

### ehild

14. Mar 19, 2016

### frillybob

I assume all of it because it says it doesn't slip.

15. Mar 19, 2016

### ehild

16. Mar 19, 2016

### frillybob

0.690N?

Not trying to irritate you or anything. It's my first time on this board.

17. Mar 19, 2016

### ehild

The torque is not force.

18. Mar 19, 2016

### frillybob

F = M*A
2.94N = 0.3Kg*9.8m/s/s?

19. Mar 19, 2016

### SteamKing

Staff Emeritus
Check your arithmetic. You calculated an incorrect value for I.

20. Mar 19, 2016

### ehild

I leave you now, I have to sleep. Read your notes or handbook or the page you cited about the torque. You might remember that you get it multiplying the force with its lever arm.
http://hyperphysics.phy-astr.gsu.edu/hbase/torq2.html

21. Mar 19, 2016

### ehild

The problem says that T is force, 0.690 N.
If it was torque, about what point is it?

22. Mar 19, 2016

### rcgldr

I'm considering the axis of the spool as the center of rotation for the net torque. If viewing this as an instant in time, then the point of contact with the surface could also be used, in which case the ground force doesn't matter, so the only torque is related to the tension in the string times the distance above the surface, but the moment of inertia about the point of contact becomes more complicated. The problem statement defines the moment of inertia about the axis of the spool, which is why I chose that as the reference point for the torques.

Back to the original question, note that the linear acceleration equals the net linear force divided by mass. The net linear force equals the tension to the right minus the surface force to the left. The angular acceleration equals the linear acceleration divided by the radius of the end caps.

Last edited: Mar 19, 2016
23. Mar 19, 2016

### haruspex

24. Mar 19, 2016

### frillybob

Thanks for that. Could you happen to explain your reasoning to the second paragraph a bit more? Could you also explain how to find it. (You can use different numbers than mine, I work much better with numbers)

25. Mar 19, 2016

### haruspex

I very strongly encourage you to develop the habit of working as far as possible without numbers. It has many advantages, and you will get better at it with practice.

With regard to the present problem, I consider it simpler to take moments about the point of contact with the ground. You just need to apply the parallel axis theorem, which I assume you have met. But go with rcgldr for now and we can look at that for comparison later. Anyway, I see this is part a) only. Depending on what the other parts ask, there might be no advantage.