Question

A professor doing a lecture demonstration stands at the center of a frictionless turntable, holding 5.00 -kg masses in each hand with arms extended so that each mass is \(1.20 \mathrm{~m}\) from his centerline. A (carefully selected!) student spins the professor up to a rotational speed of \(1.00 \mathrm{rpm} .\) If he then pulls his arms in by his sides so that each mass is \(0.300 \mathrm{~m}\) from his centerline, what is his new rotation rate? Assume that his rotational inertia without the masses is \(2.80 \mathrm{~kg} \mathrm{~m} / \mathrm{s}\), and neglect the effect on the rotational inertia of the position of his arms, since their mass is small compared to the mass of the body.

Short answer

Answer: The new rotation rate of the spinning professor is approximately 4.64 rpm.

Step-by-step solution

Calculate the initial rotational inertia of the system

To find the initial rotational inertia, We need to first calculate the rotational inertia of the two masses being held by the professor, and then add it to the professor's own rotational inertia. Since the masses are held at a distance of \(1.20 \mathrm{~m}\) from the professor's centerline, we can use the formula for the rotational inertia of a point mass: \(I = m * r^2\) So the rotational inertia of each mass is: \(I_{mass} = 5 \times 1.20^2 = 7.2 \mathrm{~kg} \mathrm{~m^2}\) Now, we add the rotational inertia of the two masses to the professor's rotational inertia (given as \(2.80 \mathrm{~kg} \mathrm{~m^2}\)): \(I_{total} = I_{professor} + 2 * I_{mass} = 2.80 + 14.40 = 17.20 \mathrm{~kg} \mathrm{~m^2}\)

Calculate the initial angular momentum of the system

Now we can calculate the initial angular momentum of the system. We will first convert the given rotational speed of \(1.00 \mathrm{rpm}\) to radians per second: \(\omega_{initial} = \frac{2 * \pi * 1}{60} \approx 0.1047 \mathrm{rad/s}\) Now we can find the initial angular momentum using the formula: \(L_{initial} = I_{total} \times \omega_{initial} = 17.20 \times 0.1047 \approx 1.80 \mathrm{~kg} \mathrm{~m^2/s}\)

Calculate the final rotational inertia of the system

Now we need to calculate the final rotational inertia of the system after the professor pulls the masses in. The new distance from the centerline is \(0.300 \mathrm{~m}\), so the rotational inertia of each mass is now: \(I'_{mass} = 5 * 0.300^2 = 0.45 \mathrm{~kg} \mathrm{~m^2}\) Now, we add the rotational inertia of the two masses to the professor's rotational inertia: \(I'_{total} = I_{professor} + 2 * I'_{mass} = 2.80 + 0.90 = 3.70 \mathrm{~kg} \mathrm{~m^2}\)

Calculate the final angular momentum and rotation rate

Since the angular momentum is conserved, the final angular momentum (\(L_{final}\)) will be equal to the initial angular momentum: \(L_{final} = 1.80 \mathrm{~kg} \mathrm{~m^2/s}\) Now we can find the final rotation rate by dividing the final angular momentum by the final rotational inertia: \(\omega_{final} = \frac{L_{final}}{I'_{total}} = \frac{1.80}{3.70} \approx 0.486 \mathrm{rad/s}\) Finally, we can convert the final rotation rate back to rotations per minute: \(\omega_{final} = \frac{0.486 * 60}{2 * \pi} \approx 4.64 \mathrm{rpm}\) So the professor's new rotation rate is approximately \(4.64 \mathrm{rpm}\).

Key concepts

Angular Momentum Conservation

To grasp the principle of angular momentum conservation, envision a figure skater spinning. As they pull their arms in, they spin faster, and when they extend their arms, they slow down. This occurs because angular momentum, the rotational equivalent of linear momentum, is conserved in the absence of external torques. In our example, the professor pulling in the weights is similar to the skater drawing in their arms.

Angular momentum, denoted by the symbol \( L \), is the product of an object's moment of inertia \( I \) and its angular velocity \( \omega \). It's given by the equation \( L = I \times \omega \). When the professor changes the position of the weights, the system's angular momentum remains constant because no external forces are acting on it. Therefore, the initial and final angular momenta are equal, and we can write \( L_{initial} = L_{final} \). This allows us to connect the initial and final rotation rates once we've calculated the moments of inertia for each configuration, as shown in the exercise solution.

Rotational Kinematics

Understanding Angular Velocity

Rotational kinematics deals with the motion of rotating bodies. In the context of the exercise, when the professor's rotation rate changes, we are observing rotational kinematics in action. Angular velocity, denoted by \( \omega \), is the rate at which an object rotates, analogous to linear velocity in translational motion.

To transition between rotations per minute (rpm) and the standard radians per second (rad/s), we use the fact that there are \( 2\pi \) radians in one rotation and 60 seconds in one minute. This allows us to accurately determine the professor's rotation rates before and after pulling the weights closer to his body.

Moment of Inertia

Rotational Inertia and Point Masses

The moment of inertia, also known as rotational inertia, is the rotational equivalent of mass in linear motion. It measures an object's resistance to changes in its angular velocity. The greater the moment of inertia, the harder it is to change an object's rotational speed.

In the exercise, we calculate the moment of inertia for point masses located at a distance from the axis of rotation using the equation \( I = m \times r^2 \), where \( m \) is the mass and \( r \) is the distance from the rotational axis. By altering the distance \( r \), the professor changes the total moment of inertia of the system. When the weights are closer, the moment of inertia decreases, and according to angular momentum conservation, his rotation rate must increase, which is exactly what the exercise demonstrates.