A hollow sphere of radius 0.250 m, with rotational inertia I = 0.0493 kg·m2 abou

Question

A hollow sphere of radius 0.250 m, with rotational inertia I = 0.0493 kg·m2 about a line through its center of mass, rolls without slipping up a surface inclined at 34.0° to the horizontal. At a certain initial position, the sphere's total kinetic energy is 11.0 J. (a) How much of this initial kinetic energy is rotational? (b) What is the speed of the center of mass of the sphere at the initial position? When the sphere has moved 0.920 m up the incline from its initial position, what are (c) its total kinetic energy and (d) the speed of its center of mass?

Explanation / Answer

Given that
radius R = 0.25 m
I = 0.0493 kg .m2
=34 degree
total kinetic energy is K = 11J

a)
The initial kinetic energy is K = (1/2)I^2 +(1/2)mU2
= (1/2)I( U/R)2 + (1/2)mU2
= (1/2)(2/3)mR2* U2 / R2 +(1/2)mU2
= (5/6)mU2
==> mU2 = 6* 11 / 5
=13.2 J
Then the rotationalkinetic energy is (1/2)I^2 = K -(1/2)mU2
=11 J - ( 13.2/2)
=4.4 J

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