Question

A think stick 1.5m long is pivoted about one end. The stick is held in a horizontal position and released from rest. Neglecting friction, use conservation of energy to determine the stick's angular velocity when it is vertical. (Hint: When determining the change in gravitational potential energy of the stick, you must consider the change in position of the stick's centre of mass)

Answer 1

Since stick is at rest initially , therefore
Initial energy = 0
Now when stick becomes vertical then
Final energy = potential energy + Kinetic energy
= mg(-L/2) + (1/2)Iw²
where I is moment of inertia of stick = mL²/3
L is length of stick
m is mass of stick
w is angular velocity
Now taking energy conservation
mg(-L/2) + (1/2)Iw² = 0
(1/2)Iw² = mg(L/2)
(1/2)(mL²/3)w² = mg(L/2)
$w^{2} = \frac{3g}{L}$
On putting the value we get
$w^{2} = \frac{3*9.81}{1.5}$
w = 4.429 rad/s