Question

A thin rod of length 0.50 m and mass 110 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 3.04 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

m = mass of rod = 110 g = 0.11 kg

L = length of rod = 0.50 m

about one end , moment of inertia of rod is given as ::

I = mL² /3

I = (0.11) (0.5)² /3

I = 0.0092 kgm²

w = angular speed = 3.04 rad/s

a)

rod's rotational kinetic energy is given as ::

RKE = (0.5)Iw²

RKE = (0.5)(0.0092) (3.04)²

RKE = 0.0425

b)

let the rod rise by height ''h''

Using conservation of energy ::

potential energy = Rotational kinetic energy

mgh = 0.0425

(0.11) (9.8) h = 0.0425

h = 0.039 m