# ROTATION:A solid ball of mass 2.6 kg and diameter 15cm is rotating about its axis at 80 rev/min. A. What is its kinetic energy? Answer in units of mJ B. If an additional 1 J of energy are supplied...

ROTATION:

A solid ball of mass 2.6 kg and diameter 15cm is rotating about its axis at 80 rev/min.

A. What is its **kinetic energy**? **Answer in units of mJ**

B. If an additional **1 J of energy** are supplied to the rotational energy, what is the new angular speed of the ball? **Answer in units of rev/min**

### 1 Answer | Add Yours

This needs the basic understanding of rotational dynamics.

Kinetic energy of a rotating mass is

KE = (I * w^2)/2

where I is the moment of inertia and w (omega) is the rotational velocity in radians per second

Now for a sphere with uniform density rotating around its axis I is,

I = (2MR^2)/5

M -mass, R -Radius

For this case,

I = (2x2.6kgx(0.15m)^2)/5 = 0.0234 kgm^2

w = 80 rpm = (80 x 2xpi)/60 radians per second

w = 8.3776 rads^-1

Therefore the kinetic energy is KE

KE = 0.5 x 0.0234 kgm^2 x 8.3776^2 s-2

**KE = 0.82115 J = 821.15 mJ**

If additonal 1Jof kinetic enrgy is inserted, the new kinetic enrgy is 1.82115 J

Since moment of inertia doesnt change, KE is directly proportionate to w^2

therefore,

(1.82115/0.82115) = (w^2)/(80^2)

w^2 = 14193.94751

**w = 119.138 rev per min**

(I used rev per min instead of rads^-1, because the units cancel out)