A hollow sphere of radius 0.250 m, with rotational inertia I = 0.0541 kg·m 2 abo

Question

A hollow sphere of radius 0.250 m, with rotational inertia I = 0.0541 kg·m2 about a line through its center of mass, rolls without slipping up a surface inclined at 13.5° to the horizontal. At a certain initial position, the sphere's total kinetic energy is 12.0 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 0.700 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

if not slipping

v=wr

total KE=0.5*mv^2+0.5*Iw^2=0.5mv^2+(0.5*(2/3)mv^2)=0.5mv^2+(1/3)mv^2

so rotational KE=12*1/3/(1/2 +1/3)=4.8 J

I=2/3mr^2=0.0541

m=1.2984 kg

total KE=(5/6)mv^2=12

we got v=3.33 m/s at inital point

(c) potential energy=mghsin(13.5)=2.07 J

KE=12-PE=9.92 J

(5/6)mv^2=9.92

we got v=3.02 m/s

kinetic energy=(5/6)*mv^2=8.53 J

v=2.8 m/s

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