A bar of mass $M$ and length $L$ is in pure translatory motion and its centre of mass has velocity $V$ . It collides and sticks to a second identical bar which is initially at rest. (Assume that it becomes one composite bar of length $2 L$ ). The angular velocity of the composite bar after collision will be :

Question

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Answer 1

Since there is no external torque, $L$ will be conserved.
$L_{i} = M V L / 2$
$L_{f} = I ω$
Since both rods stick together it becomes a rod of mass $2 M$ and length $2 L$
Hence, $L_{f} = I ω = (2 M) \frac{( 2 L ) ^{2}}{1 2} ω$
Equating $L_{i} = L_{f}$
$\Rightarrow M V L / 2 = \frac{2}{3} M L^{2} ω$
$\Rightarrow ω = \frac{3}{4} \frac{V}{L}$
The rod which has the velocity is sticking to the stationary rod which is at the top. After sticking the inertia of the stationary rod will make the whole rod to rotate in counterclockwise direction. Hence A and C is correct.

Answer 2

Since there is no external torque,

L

will be conserved.

L_{i} = M V L / 2

L_{f} = I ω

Since both rods stick together it becomes a rod of mass

2 M

and length

2 L

Hence,

L_{f} = I ω = (2 M) \frac{( 2 L ) ^{2}}{1 2} ω

Equating

L_{i} = L_{f}

\Rightarrow M V L / 2 = \frac{2}{3} M L^{2} ω

\Rightarrow ω = \frac{3}{4} \frac{V}{L}

The rod which has the velocity is sticking to the stationary rod which is at the top. After sticking the inertia of the stationary rod will make the whole rod to rotate in counterclockwise direction. Hence A and C is correct.

$\frac{3}{4} \frac{V}{L}$

$\frac{4}{3} \frac{V}{L}$

Counterclockwise

Clockwise

Revise with Concepts

Collision of Two Rigid Bodies

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Elastic Collision of Point Masses With Rigid Bodies

A bar of mass M and length L is in pure translatory motion and its centre of mass has velocity V. It collides and sticks to a second identical bar which is initially at rest. (Assume that it becomes one composite bar of length 2L). The angular velocity of the composite bar after collision will be :

43​LV​

34​LV​