Question

A thin uniform rod (length \(=1.00 \mathrm{~m},\) mass \(=2.00 \mathrm{~kg})\) is pivoted about a horizontal frictionless pin through one of its ends. The moment of inertia of the rod through this axis is \(\frac{1}{3} m L^{2} .\) The rod is released when it is \(60.0^{\circ}\) below the horizontal. What is the angular acceleration of the rod at the instant it is released?

Short answer

The angular acceleration of the rod at the instant it is released is approximately -12.75 rad/s².

Step-by-step solution

Draw the free body diagram and identify the forces

The rod is pivoted at one end, so the forces acting on it are: - gravitational force \(\vec{W}\) acting on the center of mass of the rod (which is at the middle of the rod), - and the normal force \(\vec{N}\) acting at the pivot where the rod is supported.

Apply Newton's second law in its angular form

Newton's second law in its angular form states that the net torque on an object is equal to the product of its moment of inertia and its angular acceleration: \(\tau_{net} = I \alpha\) The angle \(\theta\) between the vertical line through the pivot and the rod is \(60.0^{\circ}\). We can express the torque due to the gravitational force \(\vec{W}\) as: \(\tau_{W} = -W \cdot \frac{L}{2} \cdot \sin\theta\) The torque due to the normal force \(\vec{N}\) is zero because this force is acting through the pivot, which is the point about which we are calculating the torque. Therefore, the net torque on the rod is: \(\tau_{net} = \tau_{W}\)

Determine the angular acceleration using the moment of inertia and the net torque

We are given the moment of inertia of the rod, \(I = \frac{1}{3} mL^2\). We can now calculate the angular acceleration \(\alpha\): \(\tau_{net} = I \alpha\) Substituting the net torque and the moment of inertia: \(-W \cdot \frac{L}{2} \cdot \sin\theta = \frac{1}{3} mL^2 \alpha\) Our goal is to find the angular acceleration \(\alpha\). To do this, first, we will find the gravitational force \(W = mg\), where \(m\) is the mass of the rod and \(g\) is the acceleration due to gravity (approximately \(9.81 \mathrm{~m/s^2}\)). Then, we can solve for \(\alpha\): \(-mg\cdot\frac{L}{2}\cdot\sin\theta=\frac{1}{3}mL^2\alpha\) Divide both sides by \(mL\): \(-\frac{g}{2}\cdot\sin\theta=\frac{1}{3}L\alpha\) Now we can find \(\alpha\): \(\alpha = \frac{-3g\sin\theta}{2L}\) Plug in the values (\(g = 9.81 \mathrm{~m/s^{2}}\), \(L = 1.00 \mathrm{~m}\) and \(\theta = 60.0^{\circ}\)): \(\alpha = \frac{-3(9.81) \sin(60.0)}{2(1.00)}\) After evaluating this expression, we have: \(\alpha \approx -12.75 \, \mathrm{rad/s^{2}}\) The angular acceleration of the rod at the instant it is released is approximately \(-12.75 \, \mathrm{rad/s^{2}}\).

Key concepts

Moment of Inertia

The moment of inertia is a measure of an object's resistance to changes in its rotational motion. Much like mass in linear motion, the moment of inertia plays a central role in rotational dynamics. For a rod pivoted about one end, the moment of inertia is calculated taking into account the distribution of the rod’s mass with respect to the pivot point.

In the exercise, the rod’s moment of inertia is given as \(\frac{1}{3}mL^2\), which is derived from a general formula for a rod pivoted at the end, \(I = \frac{1}{3}ML^2\), where \(M\) is the mass and \(L\) is the length of the rod. This relationship indicates that the distribution of mass affects how easily the rod will start spinning or stop once it's in motion.

Torque

Torque is a measure of the force causing an object to rotate about an axis. If you think of it as the rotational equivalent of force, you're on the right track. The torque exerted by a force depends on the magnitude of the force, the distance from the axis of rotation (the lever arm), and the angle at which the force is applied.

In this scenario, gravity generates a torque on the rod since it is applied at a distance (lever arm) from the pivot point. The formula for torque due to the gravitational force at an angle \(\theta\) is \(\tau = -W \cdot \frac{L}{2} \cdot \sin(\theta)\). The negative sign indicates the direction of the torque which, in this case, is opposite to the chosen positive direction of rotation.

Newton's Second Law for Rotation

While most students are familiar with Newton’s second law about the relationship between force and acceleration \(F = ma\), it also has a rotational form that links torque \(\tau\), moment of inertia \(I\), and angular acceleration \(\alpha\): \(\tau_{net} = I\alpha\).

This equation shows that the net torque on an object is proportional to the angular acceleration it experiences. In the exercise, by applying this law, we can solve for the angular acceleration of the rod when the only significant torque is due to gravitational force. The absence of any external torques besides weight simplifies our calculations and directs us towards understanding how rotational forces affect motion.

Gravitational Force

Gravitational force, denoted by \(W\) in the context of this problem, is the attractive force that the Earth exerts on an object. It is calculated by the product of the mass \(m\) of the object and the acceleration due to gravity \(g\), which on Earth's surface is approximately \(9.81 \mathrm{~m/s^2}\).

The gravitational force acts on the rod’s center of mass, creating torque as it tries to rotate the rod downwards around the pivot point. The fact that this force applies to all objects equally gives us a predictable way to solve for various parameters like the one in our exercise: the angular acceleration of a rotating body.