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Chapter 1: Q37E (page 296)

A uniform sphere with a mass of 28.0 kg and a radius of 0.380 m is rotating at a constant angular velocity about a stationary axis that lies along the diameter of the sphere. If the kinetic energy of the sphere is 236 J, what is the tangential velocity of a point on the rim of the sphere?

Short Answer

Expert verified

Thus,The tangential velocity of a point on the rim of the sphere is6.49ms.

Step by step solution

01

Step:-1 explanation

Given in the question, uniform sphere rotation about a fixed diameter.

M=28.0kgR=0.380m

Constant angular velocityω.

K=236J

02

Step:-2 calculation


We know that kinetic energy given by K=12lω2

ω=vr, plug this equation into the expression for K.

K=12lv2R2

The expression F and the moment of inertia I.

K=1225MR2v2R2=15Mv2v=5KM

Put the value K=236and M=28.0.

v=523628.0=118028.0=42.14=6.49ms

Hence , The tangential velocity of a point on the rim of the sphere is6.49ms.

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