Question

A 4.00-m long rod is hinged at one end. The rod is initially held in the horizontal position, and then released as the free end is allowed to fall. What is the angular acceleration as it is released? (The moment of inertia of a rod about one end is ML2/3.)

Answer 1

Answer:

$3.75rad/s^2$

Explanation:

Let g = 10m/s2. Gravity acting on the rod:

G = mg = 10M

Suppose the rod is uniform, then we can treat gravitational torque as gravity force acting on the center of mass, which is rod's midpoint

T = 0.5GR = 0.5*10M*4 = 20M

The moment of inertia of the rod about 1 end is

$I = \frac{ML^2}{3} = \frac{M4^2}{3} = \frac{16M}{3} kgm^2$

Then the angular acceleration as soon as it releases is

$\alpha = \frac{T}{I} = \frac{20M*3}{16M} = \frac{60}{16} = 3.75rad/s^2$