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: 2/3mr², 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*w², 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)mr²*w²=2/5 mr²w²=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=1/2Iw²=1/2*(2/3)mr²*w²=1/3 mr²w²=1/3*1*(0.08)²w²=0.144

=>w²=0.144*3/(0.08)²=67.5

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