Question

A thin rod of length 0.75 m and mass 120 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 2.76 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

a) KE = 0.0857Joules

b) height = 0.073m

Step-by-step explanation:

The rotational inertia of the rod at an axis passing through the fixed end is given by:

I = Icm + my

I= 1/12 mL^2 + m(L/2)

I= 1/12 ×0.12×(0.75)^2 + 0.12× (0.75/2)^2

I = 5.625×10^-3 + 0.01688

I = 0.0225Kg/m^-2

a) KE = 1/2 Iw^2

Where I= rotational inertia

w= angular speed at lowest point

KE = 1/2 × 0.0225 ×(2.76)^2

KE= 0.0857 Joules

b) Ki + Ui = Kf + Uf

Ki = Uf - Ui = 0

mgh = K

0.0857= mgh

h = 0.0857/( 0.120× 9.8)

h = 0.0857/ 1.176

h = 0.073m