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

In summary, we used the conservation of mechanical energy, Newton's second law for rotational motion, and the moment of inertia to find the angular acceleration of the rod when it is released from a \(60.0^{\circ}\) angle below the horizontal. The calculated angular acceleration is approximately \(12.72\,\text{rad/s}^2\).

Step-by-step solution

Calculate the moment of inertia (I) of the rod

Given that the rod is thin and uniform, its moment of inertia can be calculated using the formula \(I = \frac{1}{3}mL^2\). The mass (m) is 2kg, and the length (L) is 1m. Therefore, the moment of inertia of the rod is: \(I = \frac{1}{3}(2\,\text{kg})(1\,\text{m})^2\) \(I = \frac{2}{3}\,\text{kg}\cdot\text{m}^2\)

Calculate the initial gravitational potential energy (U)

Next, we need to find the gravitational potential energy (U) at the release position. The vertical distance (h) of the center of mass of the rod from the pivot is given by \(h = \frac{1}{2}L\sin{\theta}\), where L is the length of the rod, and \(\theta\) is the angle it makes with the horizontal. Here, \(L = 1\,\text{m}\) and \(\theta = 60.0^{\circ}\). So: \(h = \frac{1}{2}(1\,\text{m})\sin{60.0^{\circ}}\) Calculating h, we get: \(h \approx 0.433\,\text{m}\) Now the gravitational potential energy (U) is given by: \(U = mgh\) \(U = (2\,\text{kg})(9.81\,\text{m/s}^2)(0.433\,\text{m})\) \(U \approx 8.50\,\text{J}\)

Calculate the net torque acting on the rod

Now we calculate the torque acting on the rod due to gravity. The gravitational force acting on the center of mass can be resolved into two components: one perpendicular to the rod and the other parallel to the rod. The torque is given by: \(\tau = F_\perp\cdot r\) Where \(F_\perp\) is the component of the gravitational force perpendicular to the rod and r is the distance of the center of mass from the pivot point. The force perpendicular to the rod can be found as: \(F_\perp = mg \sin{\theta}\) \(F_\perp = (2\,\text{kg})(9.81\,\text{m/s}^2)\sin{60.0^{\circ}}\) \(F_\perp \approx 16.97\,\text{N}\) Now, we can find the torque: \(\tau = (16.97\,\text{N})(0.5\,\text{m})\) \(\tau \approx 8.48\,\text{N}\cdot\text{m}\)

Apply Newton's second law for rotational motion

Finally, we apply Newton's second law for rotational motion, which states that: \(\tau = I\alpha\) Where \(\tau\) is the torque, I is the moment of inertia, and \(\alpha\) is the angular acceleration. We have already calculated \(\tau\) and I, so we can solve for \(\alpha\): \(\alpha = \frac{\tau}{I}\) \(\alpha = \frac{8.48\,\text{N}\cdot\text{m}}{\frac{2}{3}\,\text{kg}\cdot\text{m}^2}\) \(\alpha \approx 12.72\,\text{rad/s}^2\) Thus, the angular acceleration of the rod at the instant it is released is approximately \(12.72\,\text{rad/s}^2\).

Key concepts

Moment of Inertia

The moment of inertia is a physical quantity that represents how difficult it is to change an object's rotation about a given axis. It depends not only on the mass of the object but also on how that mass is distributed relative to the axis of rotation. For simple geometric shapes like rods, discs, and spheres, formulas are derived to calculate the moment of inertia. In our scenario with the rod, the moment of inertia through its end is given by the formula \(I = \frac{1}{3}mL^2\).

This indicates that the mass farther from the pivot contributes more to the moment of inertia than mass close to the pivot, which aligns with the intuitive understanding that it’s harder to start or stop the rotation of an extended body as opposed to a concentrated mass. In short, the larger the moment of inertia, the slower the acceleration under the same applied torque.

Gravitational Potential Energy

Gravitational potential energy (U) is the energy held by an object because of its position relative to a gravitational field. It is calculated as \(U = mgh\), where 'm' is the mass, 'g' the acceleration due to gravity, and 'h' the height above a reference point.

For the rod in our exercise, this energy is maximum when the rod is held at an angle and decreases as the rod falls. At the instant the rod is released from rest, all of this potential energy begins converting into kinetic energy (rotational in this case), allowing us to predict the motion of the rod using principles of energy conservation. Understanding gravitational potential energy is crucial in solving problems where an object's height or position changes, especially under the action of gravity.

Torque

Torque (\(\tau\)) is a measure of the force that causes an object to rotate about an axis. It is calculated by multiplying the force applied perpendicular to the object's lever arm by the length of the lever arm (\(r\)): \(\tau = F_\perp\cdot r\).

The leverage effect of torque is why doors have handles far from the hinges and why it’s easier to loosen a bolt with a long wrench. In our problem, the perpendicular component of gravity acting at the rod’s center of mass creates a torque that causes the rod to accelerate rotationally as soon as it is released. The greater the torque, the greater the object's angular acceleration, assuming a constant moment of inertia.

Newton's Second Law for Rotational Motion

Newton's second law for rotational motion extends the concept of force and acceleration to rotational motion. It says that the net torque (\(\tau\)) on an object is equal to the moment of inertia (\(I\)) multiplied by the angular acceleration (\(\alpha\)): \(\tau = I\alpha\).

This law is analogous to Newton's second law for linear motion (\(F = ma\)), where force causes acceleration. For rotation, torque takes the role of force, the moment of inertia is akin to mass, and angular acceleration is the rotational equivalent of linear acceleration. Applying this law is fundamental in calculating how an object will rotate when subjected to certain forces, as demonstrated in the rod’s scenario, where we used this principle to find the angular acceleration.