Two beads each of mass m are fixed on a light rigid rod of length 2 l which is free to rotate in a horizontal plane. The bead on the far end is given some velocity v as shown in the figure. If K cm re

Question

Two beads each of mass m are fixed on a light rigid rod of length 2 l which is free to rotate in a horizontal plane. The bead on the far end is given some velocity v as shown in the figure. If K cm represents the kinetic energy of the centre of mass of the system and K R represents the rotational kinetic energy of the system, then what is the value of K cm K R ?

TestHub · Accepted Answer

The distance of the centre of mass from the hinge,

$r_{cm} = \frac{m l + m (2 l)}{m + m} = \frac{3}{2} l$

The angular velocity of the system,

$ω = \frac{v}{2 l}$

The velocity of the centre of mass of the system,

$v_{cm} = \frac{3 l}{2} ω = \frac{3}{4} v$

$\Rightarrow K_{cm} = \frac{1}{2} 2 m {v^{2}}_{cm} = \frac{9 m v^{2}}{16}$

The moment of inertia of the system about the hinge,

$I = m l^{2} + m {(2 l)}^{2} = 5 m l^{2}$

The rotational kinetic energy of the system,

$K_{R} = \frac{1}{2} I ω^{2} = \frac{1}{2} 5 m l^{2} {(\frac{v}{2 l})}^{2}$

$\Rightarrow K_{R} = \frac{5 m v^{2}}{8}$

$\frac{K_{cm}}{K_{R}} = \frac{9 m v^{2}}{16} \times \frac{8}{5 m v^{2}} = 0.9$

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