Rolling without slipping

1. Nov 14, 2005

Punchlinegirl

A block of a certain material begins to slide on an inclined plane when the plane is inclined to an angle of 14.57°. If a solid cyclinder is fashioned from the same material, what will be the maximum angle at which it will roll without slipping on the plane?

I really have no idea where to start on this problem since the only given is the angle. I know that the moment of inertia of a cylinder is (1/2)MR^2 if thats needed at all. Any hints?

2. Nov 14, 2005

Päällikkö

In what case does the cylinder slip (hint: friction)?
Another hint: total acceleration is the same as the tangential acceleration if there's no slipping.

(Would you happen to have the correct answer?)

Last edited: Nov 14, 2005
3. Nov 14, 2005

Kenny Lee

I'm having trouble with a similar problem.

I tried finding the coeff of friction; got an arctan14.7 ...

Then I set up the dynamics equations of the cylinder, taking the center of mass as the center of rotation; Wx - friction = ma

And then I put in no slipping condition i.e a = (angular acceleration) X R.

I substitude (angular acceleration) with the relevant expression using Newton's second law in angular form.

I solve and then I get a very unlikely value of 89 degrees. Where am I wrong.

4. Nov 14, 2005

Päällikkö

According to my calculations you should get a plain tan (not arctan).

The rest sounds, if I understood everything correctly, fine.

5. Nov 14, 2005

Kenny Lee

Oh right. Yea, tan 14.7 degrees. Thanks.
okay now i get 38 degrees... dat bout right?

6. Nov 14, 2005

Punchlinegirl

I see how to do it now. Thank you

7. Nov 15, 2005

Päällikkö

If you mean the problem in this thread (with the angle actually being 14.57), I get the same answer.

No problem .

8. Nov 15, 2005

m_sepehr82

romured question

i think that it's a romured question in which the answer is 14.57 degree because friction doesnt pertain to shape of the block or how much is the surface of that
i think the answer is that