Question

A thin rod of length 0.97 m and mass 210 g 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 4.93 rad/s. 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.

Answer #1

Given:

L = 0.97m

m = 0.21kg

w at lowest point = 4.93 rad/s

Intertia through an axis passing through the fixed end:

I = I_{CM} + mr^{2}

I_{CM} is the inertia through Centre of mass of rod =
(1/12)mL^{2}

r is the distance of fixed point from Centre of mass = L/2

I = (1/12)*(0.21)*(0.97)^{2} + (0.21)*(0.97/2)^{2}
= 0.0165 + 0.0494 = 0.0659 kg.m^{2}

Part a):Kinetic Energy K is given by

**K = (1/2)Iw ^{2} = 0.5*0.0659*(4.93)^{2} =
0.801 J**

Part b): Topmost position will arise when total kinetic energy
at lowest point converts into gravitational potential energy,

K = mgh

0.801 = 0.21*9.8*h

**h = 0.389 metres**

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