Sphere rolling down a ramp linear acceleration

  • #1

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.
 

Answers and Replies

  • #2
pmp!
Apply Newton 2nd Law twice, for linear and rotational motion. Friction is providing a net torque.
 
  • #3
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.
 
  • #4
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.
 
  • #5
pmp!
The right answer is [itex]\frac{5}{7}g\sin\theta[/itex]. Did you find it? :smile:
 
  • #6
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?
 
  • #7
pmp!
You got two equations before. Solve one to get the value of F. :smile:
 
  • #8
104
0
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.
 
Last edited:
  • #9
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?
 

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