A hollow sphere of radius 0.110 m, with rotational inertia I0.0856 kg.m2 about a

Question

A hollow sphere of radius 0.110 m, with rotational inertia I0.0856 kg.m2 about a line through its center of mass, rolls without slipping up a surface inclined at 30.6° to the horizontal. At a certain initial position, the sphere's total kinetic energy is 100 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 0.810 m up the incline from its initial position, what are (c) its total kinetic energy and (d) the speed of its center of mass? (a) Number| (b) Number (c) Number (d) Number Units Units Units Units

Explanation / Answer

A)

Total Kinetic energy

K=(1/2)mV2+(1/2)IW2 =(1/2)mV2+(1/2)(2/3)mr2*(V/r)2 =(5/6)mV2

=>mV2 =(6/5)*100 =120 J

Initial Kinetic energy

Ki,rot =(1/2)IW2=(1/2)(2/3)mr2(V/r)2=(1/3)mV2=(1/3)*120

Ki,rot =40 J

b)

Since mOment of inertia of hollow sphere

I=(2/3)mr2

0.0856=(2/3)*m*0.112

m=10.61 kg

Since Krot=(1/3)MV2

40=(1/3)*10.61*V2

V=3.36 m/s

c)

By Conservation of energy

Ki=Kf+mgh

120 =Kf+10.61*9.81*(0.81sin30.6)

Kf=77.1 J

d)

Now

(5/6)mV2=77.1

V=2.95 m/s

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