Animated Solution for Physics - Rotational Motion: A rigid massless rod of length $3l$ has two masses attached at each end as shown in the figure. The rod is pivoted at point $P$ on the horizontal axis (see figure). When released from initial horizontal position, its instantaneous angular acceleration will be

Imagine a classic playground see-saw, but with a twist. Instead of being perfectly balanced in the middle, our see-saw is pivoted off-center. On the shorter arm, we have a very heavy mass, and on the longer arm, we have a lighter mass. When we let go, which way will it tilt, and how fast will it start to spin? This is the essence of rotational dynamics, and it all boils down to a battle of torques.

Let's break down the geometry of our system. We have a rigid, massless rod of total length $3l$. It is pivoted at a point $P$.

On the left side of the pivot, at a distance of $l$, sits a heavy mass $m_1=5M_0$.

On the right side of the pivot, at a distance of $2l$, sits a lighter mass $m_2=2M_0$.

The rod is initially held perfectly horizontal. The moment we release it, gravity will pull down on both masses, attempting to rotate the rod around the pivot $P$.

Newton's Second Law for Rotation

To find out how the rod accelerates rotationally, we need the rotational equivalent of Newton's Second Law ($F=ma$). For rotating bodies, this law is written as:

Here, ${\tau }_{\text{net}}$ is the net torque (the total turning effect), $I$ is the moment of inertia (how stubborn the system is against rotating), and $\alpha$ is the angular acceleration (how quickly it starts to spin). Our goal is to find $\alpha$.

Torque is calculated as the product of force and the perpendicular distance from the pivot point ($\tau =r\times F$). Since the rod is initially horizontal, the gravitational forces act perfectly perpendicular to the rod.

Let's look at the left mass. Gravity pulls it down with a force of $5M_{0}g$. It is at a distance $l$ from the pivot. This creates a torque trying to spin the rod counter-clockwise:

Now, let's look at the right mass. Gravity pulls it down with a force of $2M_{0}g$. It is at a distance $2l$ from the pivot. This creates a torque trying to spin the rod clockwise:

The two torques are fighting each other. The left side is pulling counter-clockwise with $5M_{0}gl$, and the right side is pulling clockwise with $4M_{0}gl$. The left side wins! The net torque is the difference between them:

This net torque acts in the counter-clockwise direction.

Next, we need to figure out how difficult it is to spin this system. This is measured by the moment of inertia, $I$. For a system of point masses, the moment of inertia is the sum of the mass times the square of its distance from the pivot ($I=\sum m{r}^2$).

Since the rod itself is massless, we only need to calculate the inertia for the two masses:

Plugging in our values:

Be very careful here! The distance for the right mass is $2l$, and the entire distance must be squared.

Notice how the lighter mass on the right actually contributes more to the moment of inertia ($8M_0{l}^2$) than the heavier mass on the left ($5M_0{l}^2$). This is because the distance is squared in the inertia formula. Being further away makes a mass much harder to spin!

Finding the Angular Acceleration

We now have our net torque (${\tau }_{\text{net}}=M_{0}gl$) and our moment of inertia ($I=13M_0{l}^2$). Let's plug them back into our master equation:

To find the instantaneous angular acceleration $\alpha$, we simply divide the net torque by the moment of inertia:

The $M_0$ terms cancel out, and one $l$ cancels from the numerator and denominator, leaving us with a beautifully simple final answer:

What Happens Next?

The problem specifically asks for the instantaneous angular acceleration when released from the horizontal position. Why is this word so important?

As soon as the rod starts to rotate, it is no longer horizontal. It tilts by some angle $\theta$. When this happens, the gravitational force is no longer perfectly perpendicular to the rod. The perpendicular distance from the pivot to the line of action of the force becomes $r\cos \theta$.

This means the net torque will decrease as the rod falls, following the equation ${\tau }_{\text{net}}(\theta )=M_{0}gl\cos \theta$. Because the torque decreases, the angular acceleration $\alpha$ will also decrease over time. It is only exactly $\frac{g}{13l}$ at the very instant it is released!