Question

# A thin rod of length 0.97 m and mass 210 g is suspended freely from one...

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.

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 = ICM + mr2
ICM is the inertia through Centre of mass of rod = (1/12)mL2
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.m2

Part a):Kinetic Energy K is given by
K = (1/2)Iw2 = 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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