Worked example 8.5: Hinged rod

Question: A uniform rod of mass $m=5.3 {\rm kg}$ and length $l=1.3 {\rm m}$ rotates about a fixed frictionless pivot located at one of its ends. The rod is released from rest at an angle $\theta=35^\circ$ beneath the horizontal. What is the angular acceleration of the rod immediately after it is released?

Answer: The moment of inertia of a rod of mass $m$ and length $l$ about an axis, perpendicular to its length, which passes through one of its ends is $I= (1/3) m l^2$ (see question 8.3). Hence,

$$I = \frac{5.3\times 1.3^2}{3} = 2.986 {\rm kg m^2}.$$

The angular equation of motion of the rod is

$$I \alpha = \tau,$$

where $\alpha$ is the rod's angular acceleration, and $\tau$ is the net torque exerted on the rod. Now, the only force acting on the rod (whose line of action does not pass through the pivot) is the rod's weight, $m g$. This force acts at the centre of mass of the rod, which is situated at the rod's midpoint. The perpendicular distance $x$ between the line of action of the weight and the pivot point is simply

$$x = \frac{l}{2} \cos\theta = \frac{1.3\times \cos 35^\circ}{2} = 0.532 {\rm m}.$$

Thus, the torque acting on the rod is

$$\tau = m g x.$$

It follows that the rod's angular acceleration is written

$$\alpha = \frac{\tau}{I} = \frac{m g x}{I} = \frac{5.3\times 9.81\times 0.532}{2.986} = 9.26 {\rm rad./s^2}.$$