Question

A solid sphere with mass 1.0-kg and a 16-cm-diameter turns about its diameter at 77 rev/min.

(a) What is its kinetic energy?

J

(b) If an additional 5.0 mJ of energy are supplied to the
rotational energy, what is the new angular speed of the ball?

rev/min

Answer #1

A)Moment of inertia of solid sphere is given by:
2/3mr^{2}, where m is mass of sphere, r is radius

Here,mass = 1 kg, diameter=16 cm= 0.16 m=>radius=diameter/2=0.16/2=0.08 m

Also,rotational kinetic energy is given by:1/2 *I*w^{2},
where I=moment of inertia, w is angular velocity

Here,w=77 rev/min=77*2/60 rad/s=8.0634 rad/s

So,rotational kinetic energy=
(1/2)*(2/3)mr^{2}*w^{2}=2/5
mr^{2}w^{2}=1/3*1*(0.08)^{2}(8.0634)^{2}=0.139
J

B) if an additional 5 mJ kinetic energy is supplied, kinetic energy becomes 139+5 mJ = 144 mJ=0.144 J

Let the final angular be w rad/s.

So,final kinetic
energy=1/2Iw^{2}=1/2*(2/3)mr^{2}*w^{2}=1/3
mr^{2}w^{2}=1/3*1*(0.08)^{2}w^{2}=0.144

=>w^{2}=0.144*3/(0.08)^{2}=67.5

=>w=8.22 rad/sec=8.22*60/(2)=78.50 rev/min

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