Problem 34 - College Physics

Question

A large wooden turntable in the shape of a flat disk has a radius of 2.00 \(\mathrm{m}\) and a total mass of 120 \(\mathrm{kg}\) . The turntable is initially rotating at 3.00 \(\mathrm{rad} / \mathrm{s}\) about a vertical axis through its center. Suddenly, a 70.0 kg parachutist makes a soft landing on the turntable at a point on its outer edge. Find the angular speed of the turntable after the parachutist lands. (Assume that you can treat the parachutist as a particle.)

Step-by-Step Solution

Answer

The angular speed of the turntable after the parachutist lands is approximately 1.385 rad/s.

Step 1: Determine the Moment of Inertia of the Turntable Before Landing

The moment of inertia of a uniform flat disk about its central axis is given by \( I = \frac{1}{2} m r^2 \), where \( m \) is the mass of the disk and \( r \) is its radius. Here, \( m = 120 \, \mathrm{kg} \) and \( r = 2.00 \, \mathrm{m} \). Thus:\[ I_{disk} = \frac{1}{2} \times 120 \, \mathrm{kg} \times (2.00 \, \mathrm{m})^2 = 240 \, \mathrm{kg} \cdot \mathrm{m}^2 \]

Step 2: Determine the Moment of Inertia of the Parachutist

Since the parachutist lands at the edge of the disk, treat the parachutist as a point mass at radius \( r = 2.00 \, \mathrm{m} \) from the center. The moment of inertia \( I \) for a point mass is given by \( I = m r^2 \). For the parachutist:\[ I_{parachutist} = 70.0 \, \mathrm{kg} \times (2.00 \, \mathrm{m})^2 = 280 \, \mathrm{kg} \cdot \mathrm{m}^2 \]

Step 3: Calculate the Total Initial Angular Momentum

The initial angular momentum is given by \( L = I \omega \), where \( I \) is the moment of inertia of the system and \( \omega \) is the angular speed. Initially, only the disk is rotating with \( \omega = 3.00 \, \mathrm{rad/s} \). Therefore:\[ L_{initial} = I_{disk} \times \omega = 240 \, \mathrm{kg} \cdot \mathrm{m}^2 \times 3.00 \, \mathrm{rad/s} = 720 \, \mathrm{kg} \cdot \mathrm{m}^2/\mathrm{s} \]

Step 4: Determine the Total Moment of Inertia After Landing

After the parachutist lands on the turntable, the total moment of inertia is the sum of the turntable and the parachutist:\[ I_{total} = I_{disk} + I_{parachutist} = 240 \, \mathrm{kg} \cdot \mathrm{m}^2 + 280 \, \mathrm{kg} \cdot \mathrm{m}^2 = 520 \, \mathrm{kg} \cdot \mathrm{m}^2 \]

Step 5: Use Angular Momentum Conservation to Find Final Angular Speed

Angular momentum is conserved since there are no external torques acting on the system. Thus, \( L_{initial} = L_{final} \). The final angular momentum is \( I_{total} \times \omega_{final} \).\[ 720 \, \mathrm{kg} \cdot \mathrm{m}^2/\mathrm{s} = 520 \, \mathrm{kg} \cdot \mathrm{m}^2 \times \omega_{final} \]Solve for \( \omega_{final} \):\[ \omega_{final} = \frac{720}{520} = 1.385 \, \mathrm{rad/s} \]

Key Concepts

Moment of Inertia

When dealing with rotational motion, one important concept is the moment of inertia. Think of it like the rotational equivalent of mass in linear motion. It tells us how difficult it is to change the rotational state of an object. The larger the moment of inertia, the harder it is to spin an object. It depends on both the mass of the object and how that mass is distributed with respect to the axis of rotation.

For a flat, uniform disk like the turntable in our problem, the formula for the moment of inertia is given by \( I = \frac{1}{2} m r^2 \). Here \( m \) represents the mass of the disk, and \( r \) is its radius. In our case, the disk’s moment of inertia was calculated to be \( 240 \, \mathrm{kg} \cdot \mathrm{m}^2 \). Then, for the parachutist treated as a point mass on the edge of the disk, the moment of inertia is simply \( I = m r^2 \). This resulted in \( 280 \, \mathrm{kg} \cdot \mathrm{m}^2 \). This shows how both objects contribute to the total rotational inertia of the system after the parachutist lands.

Conservation of Angular Momentum

Angular momentum is a key concept in rotational dynamics. It is a conserved quantity, similar to linear momentum in traditional motion. This means that if no external torque is applied to a system, its angular momentum remains constant, even if its internal dynamics change.

In this exercise, when the parachutist lands on the turntable, even though the mass and distribution change, the total angular momentum of the system is preserved since there's no external force out there altering it. Initially, only the disk is spinning, but once the parachutist lands, both together share what was the initial angular momentum. The final angular speed drops to \( 1.385 \, \mathrm{rad/s} \) as calculated, because the total moment of inertia has increased.

This is a great demonstration of the conservation principle at work: even as the configuration of a system changes, its intrinsic angular momentum remains unchanged.

Physics Problem Solving

Solving physics problems, like this one, requires breaking them into steps and understanding the core concepts involved. In this problem, the steps focused on understanding how to calculate the moment of inertia, set up equations for angular momentum, and apply conservation principles.

Breaking down the problem:

• You first find out the turntable's moment of inertia alone using its formula for a disk.
• Next, account for the parachutist's moment of inertia given their mass at the rim of the disk.
• Combine these values to get the total system moment of inertia after landing.
• Use the conservation of angular momentum to find the system's final angular speed.

Understanding the underlying principles and then having a step-by-step process can simplify what may at first seem complex.

Rotational Motion

Rotational motion involves objects that are rotating around an axis. In physics, this kind of motion is critical because many objects rotate, from simple wheels to complex systems like planets.

Key quantities in rotational motion include angular velocity, moment of inertia, and angular momentum, all of which are featured in this exercise.

For instance, the turntable initially spinning at \( 3 \, \mathrm{rad/s} \) exemplifies angular velocity, which measures how fast the object rotates. The parachutist landing alters the distribution and hence the moment of inertia, leading to a new angular velocity, due to the preserved angular momentum. This showcases how interconnected these aspects are in rotational dynamics.