Question

A hollow sphere of radius 0.120 m, with rotational inertia I = 0.0612 kg·m² about a line through its center of mass, rolls without slipping up a surface inclined at 33.8° to the horizontal. At a certain initial position, the sphere's total kinetic energy is 42.0 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 0.610 m up the incline from its initial position, what are (c) its total kinetic energy and (d) the speed of its center of mass?

Answer 1

(a) Total kinetic energy = translational + rotational

Moment of inertia of hollow sphere = 2mr² / 3

now, total kinetic energy

E = 1/2mv² + 1/2Iw²

w = vr

E = 1/2mv² + 1/2 * 2mr² / 3 * v² / r²

E =  1/2mv² + 1/3mv²

E = 5/6mv²

42 = 5/6mv²

mv² = 50.4

so,

rotational kinetic energy = 1/3mv²

rotational kinetic energy =   50.4/3

rotational kinetic energy = 16.8 J

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(b) 0.0612 = 2mr² / 3

solve for m

0.0612 = 2m (0.12)² / 3

m = 6.375 Kg

so

5/6mv² = 42

v = 2.811 m/s

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When the sphere has moved 0.610 m up

mgh = 6.375 *9.8*0.61* sin33.8

mgh = 21.2 J

so,

total K.E = 42 - 21.2

total K.E = 20.7 J

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(d) v = sqrt (6*20.7 / 5*6.375)

v = 1.97 m/s