## Chapter 20

## Q. 20.1

A rigid body consists of two particles P_{1} and P_{2} of masses m_{1} and m_{2} attached to a light rod AB as shown in figure 20.4. AP_{1} = r_{1} and AP_{2} = r_{2} and P_{2} is at B.

Find the kinetic energy of the body when it rotates with angular speed ω about an axis perpendicular to the rod

i) through A ii) through B.

## Step-by-Step

## Verified Solution

i) When the rod rotates with angular speed ω about an axis through A perpendicular to the rod, each particle moves in a circle, centre A.

P_{1} has speed v_{1} = r_{1}ω and P_{2} has speed v_{2} = r_{2}ω . The total kinetic energy is, therefore,

\frac{1}{2}m_{1}v_{1}^{2} = \frac{1}{2}m_{2}v_{2}^{2} = \frac{1}{2}m_{1}r_{1}^{2}ω^{2} + \frac{1}{2}m_{2}r_{2}^{2}ω^{2}

= \frac{1}{2}(m_{1}r_{1}^{2} + m_{2}r_{2}^{2})ω^{2}.

ii) When the axis is through the end B, the particle P_{2} at B does not move so has no kinetic energy. The particle P_{1} now moves in a circle of radius ( r_{2} – r_{1} ), so the total kinetic energy is now \frac{1}{2}m_{1}(r_{2} – r_{1})^{2}ω^{2}