# A thin rod of length 0.75 m and mass 0.42 kg is suspended freely from one end. It is pulled to...

## Question:

A thin rod of length 0.75 m and mass 0.42 kg is suspended freely from one end. It is pulled to one side and then allowed to swing like a pendulum, passing through its lowest position with angular speed. Neglecting friction and air resistance, find:

a) the rod's kinetic energy at its lowest position and

b) how far above that position the center of mass rises.

## Rotation Kinetic energy:

The kinetic energy possessed by a body in rotational motion depends on moment of inertial in the similar manner as the mass does in linear motion. The center of mass rises to the height where its final velocity is zero. The maximum height is given by the work energy theorem applying for the rotational kinetic energy and work is done by gravity.

Given:

Length of rod,

{eq}L=0.75\ m \\ {/eq}

Mass of rod, {eq}m=0.42\ kg {/eq}

Angular speed {eq}\omega =4\ rad/s {/eq}

Moment of inertia of rod about its end, {eq}I= \dfrac{1}{3}ML^2 \\ I= \dfrac{1}{3} \times 0.42 \times 0.75^2 \\ I=0.07875\ kg.m^2 {/eq}

a)

Rod's kinetic energy at the lowest position {eq}=\dfrac{1}{2}I \omega^2 \\ =0.5 \times 0.07875 \times 4^2 \\ = 0.63\ J {/eq}

b)

The center of mass rises up to a position where kinetic energy of rod fully converted into potential energy. If 'h' be the height of rod

{eq}mgh=0.63 \\ 0.42 \times 9.8 h=0.63 \\ h=0.153\ m {/eq} 