Rolling without slipping down a hill

[IWuvTeTwis]

Homework Statement

A basketball rolls without slipping down an incline of angle θ. The coefficient of static friction is μs. Model the ball as a thin spherical shell. (Use any variable or symbol stated above along with the following as necessary: m for the mass of the ball and g.) ]

a) Find the acceleration of the center of mass of the ball

I made this picture of the forces for this situation: http://img193.imageshack.us/img193/43/balldownhill.png

Homework Equations

F = ma
τ = Iα
τ = rxF

The Attempt at a Solution

I found that
mgsinθ + μsmgcosθ = ma
solving for a we get a = gsinθ + μsgcosθ which doesn't seem to be the right answer

additionally from τ = Iα, I found Fs*r = Iα since the only force causing torque is Fs
This equals μsmgcosθ * r = mr^2*a/r
which simplifies down to a = μsgcosθ

obviously neither of these are correct so I'm confused as to where I went wrong

 

[Doc Al]

I found that
mgsinθ + μsmgcosθ = ma

(1) Do not assume that the static friction is at its maximum value.
(2) Do you think that the presence of friction increases the acceleration?

 

[IWuvTeTwis]

would the translational form of Newton's second law be mgsinθ - Fs = ma then?
And if so, I can use Newton's second law for rotation to find that Fs*r = mR^2*a/r getting that ma = Fs
Plugging this into the translational form of the equation I would obtain a = 1/2gsinθ. Is that the answer then?

 

[Doc Al]

would the translational form of Newton's second law be mgsinθ - Fs = ma then?

Exactly.

And if so, I can use Newton's second law for rotation to find that Fs*r = mR^2*a/r getting that ma = Fs

Right idea, but wrong rotational inertia. (It's a spherical shell, not a cylindrical shell.)

Plugging this into the translational form of the equation I would obtain a = 1/2gsinθ. Is that the answer then?

Redo your rotation equation and you'll have it.

 

[mistasong]

Would the frictional force acting on the ball then be 2/3*ma?

 

[Doc Al]

Would the frictional force acting on the ball then be 2/3*ma?

Yes.

 

[mistasong]

I am trying to figure out what the maximum incline can be for the ball to roll without slipping. I know that the nonslip condition means a = rα but I do not understand how to connect it with the problem.

 

[Doc Al]

I am trying to figure out what the maximum incline can be for the ball to roll without slipping. I know that the nonslip condition means a = rα but I do not understand how to connect it with the problem.

The nonslip condition is part of it. What you want to solve for is the angle that makes the friction force equal to its maximum value. That will depend on the coefficient of friction, of course.

 

[mistasong]

Okay so I find that a = 3/5gsinθ plugging that into f = 2/3ma i get f = 2/5mgsinθ. Because static friction is at its maximum value, I set the equation as μsmgcosθ = 2/5mgsinθ and solving for θ gets me 5/2μs = tanθ. Is this correct?

 

[Doc Al]

Okay so I find that a = 3/5gsinθ plugging that into f = 2/3ma i get f = 2/5mgsinθ. Because static friction is at its maximum value, I set the equation as μsmgcosθ = 2/5mgsinθ and solving for θ gets me 5/2μs = tanθ. Is this correct?

Yes. That looks good.