Question

When a stick of mass m and length | = 0.6 m is released from vertical position (as shown in figure), then what is the velocity of its free end when it strikes the ground?​

Answer 1

Answer:

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Answer 2

Answer:

Given:

Stick of mass m , length = 0.6 m is released from vertical position.

To find:

Velocity of free end when it strikes the ground.

Concept:

The basic concept to be used is that : The whole potential energy at vertical position will be converted to Rotational Kinetic Energy in the horizontal position.

Calculation:

Let the free end of rod be v , we know that centre of mass of a rod is at its midpoint. so gravitational force will act at l = 0.3 m.

$PE \: = \: KE$

$= > mg(0.3) = \dfrac{1}{2} I {\omega}^{2}$

$= > mg(0.3) = \dfrac{1}{2} \times ( \dfrac{m {l}^{2} }{3} ) {\omega}^{2}$

We know that ω = v/l , we can say :

$= > \cancel mg(0.3) = \dfrac{1}{2} \times ( \dfrac{ \cancel m {l}^{2} }{3} ) \times { (\dfrac{v}{l}) }^{2}$

Cancelling all other terms :

$= > {v}^{2} = 3 \times 3 \times 2$

$= > v = 3 \sqrt{2} \: m {s}^{ - 1}$

So final answer :

$\boxed{ \red{ \huge{ \bold{v = 3 \sqrt{2} \: m {s}^{ - 1} }}}}$