Sphere rolling down a ramp linear acceleration

[velvetymoogle]

Homework Statement

"A solid sphere of radius R and mass M is initially at rest at the top of a ramp. The lowest point of the sphere is a vertical h above the base of the ramp. It is released and rolls without slipping down the ramp. Determine the linear acceleration while the sphere is anywhere on the ramp.

M (mass), R (radius), h (height), g (gravity), theta

Homework Equations

conservation of momentum
I = 2/5MR^2
w = v/r

The Attempt at a Solution

I ended up finding the linear velocity anywhere on the ramp to be square root of 10gh/7. How would I be able to use that though to find acceleration? I seriously don't know what else to do.

 

[pmp!]

Apply Newton 2nd Law twice, for linear and rotational motion. Friction is providing a net torque.

 

[velvetymoogle]

That requires that I find the frictional force, but that's the next question so there has to be some way I can do it without knowing the friction force.

 

[pmp!]

You don't need to calculate the friction force. Just call it F and get the value from one of the equations and substitute in the other equation.

 

[pmp!]

The right answer is $\frac{5}{7}g\sin\theta$. Did you find it?

 

[velvetymoogle]

Yes, I did. Thanks for your help. Would you be able to explain how I would get the frictional force then? Is it just F = m times that or is it -mgsin(theta) because it's on a ramp?

 

[pmp!]

You got two equations before. Solve one to get the value of F.

 

[AngeloG]

pmp! is correct.

It'd be helpful to draw a force body diagram to really grasp the idea of this =).

There's a normal force, a force of a friction, a force of gravity pulling it down in the x and y direction. There's an incline on the ramp (theta).

We know in the Y direction it's not accelerating; thus we can set that summination in the Y axis to 0. However in the X axis we do have a mass that's accelerating; thus we can set that summination in the X direction to mass * acceleration.

After you setup your forces you can then figure it all out mathematically.

 

[velvetymoogle]

Oooh okay. That really helps. (I didn't originally use force equations to find the linear acceleration). I used vf^2 = vi^2 + 2ad. Is the correct answer:
(2mgsin(theta))/7?