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## 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. ## 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 $\frac{1}{2}m_{1}(r_{2} – r_{1})^{2}ω^{2}$