A hollow sphere of radius 0.150 m, with rotational inertia I = 0.0443 kg·m2 abou

Question

A hollow sphere of radius 0.150 m, with rotational inertia I = 0.0443 kg·m2 about a line through its center of mass, rolls without slipping up a surface inclined at 24.6° to the horizontal. At a certain initial position, the sphere's total kinetic energy is 20.0 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 0.510 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

part a )

Krot/Ktotal

Ktotal = 1/2*mv^2 + 1/2*Iw^2

I = 2/3mr^2

v = wr

Ktotal = 1/2*mr^2w^2 + 1/3*mr^2w^2

Krot = 1/3*mr^2w^2

Krot/K = 0.4

Krot = 0.4*K

initiaaly K = 20J

Krot = 20*0.4 = 8J

part b )

Krot = 1/3 * mr^2w^2 = 8J

I = 2/3mr^2

m = 3I/2r^2

m = 2.95 kg

w = sqrt(3*Krot/mr^2)

w = 19 rad/s

vcm = w*r = 19 * 0.15 = 2.85 m/s

part c )

Ki = Kf + mgh

Ki = 20 J

h = 0.510 *sin24.6 = 0.212m

kf = 13.86 J

part d )

in part a we found Krot is 40% of total kinetic energy

Krot = 5.54566 J

wf = sqrt(3*5.54566/0.15^2*2.95)

wf = 15.83 rad/s

vcm final = 15.83 * 0.15 = 2.375 m/s

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