First slide

Law of conservation of angular momentum

Question

A merry-go-round made of a ring-like platform of radius R and mass M, is revolving with angular speed $ω$ . A person of mass M is standing on it. At one instant, the person jumps off the round, radially away from the centre of the round (as seen from the round). The speed of the round afterwards is

Moderate

A

2 $ω$
B

$ω$
C

$\frac{ω}{2}$
D

0

Solution

As no external torque acts on the system, angular momentum should be conserved.
Hence. I $ω$ = constant …. (i)
where, I is moment of inertia of the system and $ω$ is angular
velocity of the system.
From Eq. (i), $I_{1} ω_{1} = I_{2} ω_{2}$
(where, $ω$ ₁ and $ω$ ₂ are angular velocities before and after jumping)
$\Rightarrow I ω = \frac{I}{2} \times ω_{2}$
(As mass reduced to half, hence moment of inertia also reduced to half)
$\Rightarrow ω_{2} = 2 ω$

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