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 + mr²
I_CM 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