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Rotational Sphere on Slope

  • Thread starter einstein18
  • Start date
  • #1
Ok, im trying to understand this problem on my practice exam and I can't figure out what im doing wrong.

Homework Statement


A solid sphere of mass M and radius R is released from rest on an inclined plane with an angle of θ. The coefficient of static friction for the sphere on the plane is μs. Assuming that the sphere rolls without slipping down the plane and that the static frictional force is at its maximum value, which of the following is the correct equation for the acceleration of the center of mass of the sphere?

Homework Equations


Since it is static friction and the sphere doesn't slip:
X-axis: Mgsinθ = Fstatic
Y-axis: Mgcosθ = N
Torque: Fstatic*R = I*alpha
alpha = a/R
I of solid sphere = (2/5)MR^2

The Attempt at a Solution


Simplifying the torque equation and making substituitions:
a = Fstatic*R^2 / I -> a = (μMgcosθ)R^2 / (2/5)MR^2 ->

a = μgcosθ/(2/5)

However, the answer is: a = gsinθ - μgcosθ

I see where the gsinθ comes from but I dont understand why its there if μ is static.
Also the thing that is confusing me the most, where does (2/5) go???

Any help would be greatly appreciated,
thanks in advance!
 

Answers and Replies

  • #2
I just realized that if I solve for the X-axis equation i get the answer:

mgsinθ - f = ma
a = gsinθ - μgcosθ

But this is only true if the friction were kinetic. Also where does the rotational aceleration go?
 
  • #3
Oh!!! The question asks for the equation of the acceleration of the center of mass of the sphere. Doh. I cant believe i just spent the past hour trying to figure this out...
 
  • #4
29
0
Einstein eh :P
 

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