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A uniform solid sphere rolls down an incline.

(a) What must be the incline angle if the linear acceleration of the center of the sphere is to have a magnitude of 0.10g?
(b) If a frictionless block were to slide down the incline at that angle, would its acceleration magnitude be more than, less than, or equal to 0.10g? Why?

User KitsuneYMG
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1 Answer

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Answer with Explanation:

We are given that

a.Linear acceleration,a=0.1 g

Moment of inertia of solid sphere,
I=(2)/(5)mr^2

We know that


a=(gsin\theta)/(1+(I)/(mr^2))

Using the formula


0.1g=(gsin\theta)/(1+(2mr^2)/(5mr^2))


0.1(1+(2)/(5))=sin\theta


sin\theta=0.1* (7)/(5)=0.14


\theta=sin^(-1)(0.14)=8.05^(\circ)

b.When a frictionless block were to slide down the incline at that angle then the motion of block will be linear .There is no rotation.

When the sphere is moving without rotation then, the linear speed of sphere will be larger.

When the linear speed is larger then, the linear acceleration will be larger.

Hence, the magnitude of acceleration will be more.

User Jason Tyler
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