# A ball weighing 10 kg rotates at 10 revolutions per second in a circle with radius 10 m. What is the work required to bring it 5 m closer to the center.

### 1 Answer | Add Yours

The ball weighing 10 kg is attached to a string and rotated in a circular path that has a radius of 10 m. The ball completes 10 rotations per second or the angular velocity is 10*2*pi radians/sec.

The linear velocity of the ball is v = 10*10*2*pi m/s

The kinetic energy of the ball is 10*(10*10*2*pi)^2/10

When the ball is brought closer to 5 m away from the center, the kinetic energy of the ball is 10*(10*10*2*pi)^2/5

The increase in kinetic energy is the work that is required. This is given by

10*(10*10*2*pi)^2/5 - 10*(10*10*2*pi)^2/10

=> 10*(10*10*2*pi)^2*(1/5 - 1/10)

=> 10*(10*10*2*pi)^2/10

=> (10*10*2*pi)^2

=> 4*pi^2*10^4

=> 125.6 kJ

**The work required to pull the ball closer while it rotates at the same speed is 125.6 kJ**