A sphere rolls down an inclined plane and the horizontal is 45.0 degrees. The mo

Question

A sphere rolls down an inclined plane and the horizontal is 45.0 degrees. The moment of inertia of the sphere is (2/5)mR^2), where m is its mass and R is its radius. Determine the minimum coefficient of static friction between the sphere and the plane so that the sphere rolls without slipping. Assume the coefficient of rolling friction is zero, the plane is stationary, and the force caused by the air is negligible. Hint: Use Newtons 2nd Law for translational motion and Newtons 2nd law for rotational motion

Explanation / Answer

ma = (force due to gravity along slope) - (force due to friction)
ma = mgsin - Ff [1]

torque = (moment of inertia of sphere)(angular acceleration) = (force due to friction)(radius of sphere)
I = (Ff)(r) [2]

Since the sphere doesn't slip, for [2]:
(2mr^2/5)(a/r) = (Ff)(r)
Ff = 2ma/5 [3]

Substituting Ff into [1]:
ma = mgsin - 2ma/5
a = 5gsin / 7 [4]

we know,   Ff = mgcos [5]

from 3,4 ,5

(2/7)mgsin =mgcos

= (2/7)tan

= (2/7)=0.2857

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