When a stick of mass m and length | = 0.6 m is released from vertical

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

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»

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.