Problem 45 - University Physics with Modern Physics

Question

A large wooden turntable in the shape of a flat uniform 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 near the outer edge. (a) Find the angular speed of the turntable after the parachutist lands. (Assume that you can treat the parachutist as a particle.) (b) Compute the kinetic energy of the system before and after the parachutist lands. Why are these kinetic energies not equal?

Step-by-Step Solution

Answer

(a) 1.385 rad/s; (b) 1080 J before, 498 J after; Energy is lost due to non-conservative forces.

Step 1: Identify Initial Moment of Inertia

The moment of inertia of a uniform disk rotating about an axis through its center is given by \( I_1 = \frac{1}{2} M R^2 \), where \( M \) is the mass of the disk and \( R \) is the radius. Substituting the given values, \( M = 120 \, \text{kg} \) and \( R = 2.00 \, \text{m} \), we get:\[ I_1 = \frac{1}{2} \times 120 \, \text{kg} \times (2.00 \, \text{m})^2 = 240 \, \text{kg} \cdot \text{m}^2 \].

Step 2: Calculate Initial Angular Momentum

The initial angular momentum \( L_1 \) of the turntable can be calculated using the formula \( L = I \omega \), where \( \omega \) is the angular speed. Given that \( \omega_1 = 3.00 \, \text{rad/s} \), we calculate:\[ L_1 = I_1 \omega_1 = 240 \, \text{kg} \cdot \text{m}^2 \times 3.00 \, \text{rad/s} = 720 \, \text{kg} \cdot \text{m}^2/\text{s} \].

Step 3: Compute Moment of Inertia with Parachutist

After the parachutist lands on the edge of the turntable, his moment of inertia can be included as \( I_2 = m r^2 \), where \( m = 70.0 \, \text{kg} \) and \( r = R = 2.00 \, \text{m} \). Therefore, the parachutist's moment of inertia is:\[ I_\text{parachutist} = 70.0 \, \text{kg} \times (2.00 \, \text{m})^2 = 280 \, \text{kg} \cdot \text{m}^2 \].

Step 4: Total Final Moment of Inertia

The total moment of inertia of the system after the parachutist lands is the sum of the moment of inertia of the disk and the parachutist:\[ I_\text{total} = I_1 + I_\text{parachutist} = 240 \, \text{kg} \cdot \text{m}^2 + 280 \, \text{kg} \cdot \text{m}^2 = 520 \, \text{kg} \cdot \text{m}^2 \].

Step 5: Apply Conservation of Angular Momentum

According to the conservation of angular momentum, the initial angular momentum should equal the final angular momentum. Therefore, \( L_1 = L_2 = I_\text{total} \omega_2 \), where \( \omega_2 \) is the final angular speed. Solving for \( \omega_2 \):\[ \omega_2 = \frac{L_1}{I_\text{total}} = \frac{720 \, \text{kg} \cdot \text{m}^2/\text{s}}{520 \, \text{kg} \cdot \text{m}^2} \approx 1.385 \, \text{rad/s} \].

Step 6: Kinetic Energy Before Landing

The kinetic energy of a rotating object is given by \( KE = \frac{1}{2} I \omega^2 \). For the turntable before the parachutist lands:\[ KE_1 = \frac{1}{2} \times 240 \, \text{kg} \cdot \text{m}^2 \times (3.00 \, \text{rad/s})^2 = 1080 \, \text{J} \].

Step 7: Kinetic Energy After Landing

After the parachutist lands, we calculate the new kinetic energy:\[ KE_2 = \frac{1}{2} \times 520 \, \text{kg} \cdot \text{m}^2 \times (1.385 \, \text{rad/s})^2 \approx 498 \, \text{J} \].

Step 8: Reason for Difference in Kinetic Energies

The kinetic energies before and after are not equal because the system is not closed in terms of energy; work done by non-conservative forces during the parachutist's landing results in a loss of mechanical energy, hence the reduced kinetic energy.

Key Concepts

Rotational Dynamics

Rotational dynamics deals with the motion of objects that rotate. In this context, it's about the large wooden turntable and the parachutist landing on it. Just like linear dynamics involves terms like mass and linear velocity, rotational dynamics involves 'moment of inertia' and 'angular velocity'.

Angular velocity is similar to linear velocity but applies to objects moving along a circular path. It's measured in radians per second (rad/s) and it's a key player in rotational dynamics.

Angular momentum, noted by \( L \), plays a similar role to linear momentum. It's the product of an object's moment of inertia \( I \) and its angular velocity \( \omega \). The principle of conservation of angular momentum states that if no external torque acts on a system, the total angular momentum of the system remains constant. This is what allows the angular speed of the turntable to change when the parachutist lands, but the overall angular momentum of the system to remain unchanged. This concept is crucial to solving problems involving rotating systems such as this one.

Moment of Inertia

The moment of inertia \( I \) is a measure of an object's resistance to changes in its rotation. It depends on the distribution of the mass of the object relative to the axis about which it's rotating. For instance, a turntable or disc shaped object will have a different moment of inertia compared to a parachutist due to their mass configurations.

In a formula, for a disc like the turntable, it's given by \( I = \frac{1}{2}MR^2 \), where \( M \) is the mass and \( R \) is the radius.

• The moment of inertia of the turntable is \( 240 \text{ kg} \cdot \text{m}^2 \).
• The parachutist contribution, when he is near the edge, is \( 280 \text{ kg} \cdot \text{m}^2 \).

The sum of these yields a final moment of inertia for the system of \( 520 \text{ kg} \cdot \text{m}^2 \), which significantly impacts the system's rotational speed after landing. Understanding moment of inertia aids in predicting how different objects will react under rotational forces.

Kinetic Energy in Rotational Motion

Kinetic energy in rotational motion can be thought of in much the same way as kinetic energy in linear motion, but with rotational parameters. It's given by \( KE = \frac{1}{2} I \omega^2 \), where \( I \) is the moment of inertia and \( \omega \) is the angular velocity.

Before the parachutist lands, the turntable has a higher angular velocity, resulting in a kinetic energy of \( 1080 \text{ J} \). After he lands, the total system moment of inertia increases, but the turntable's angular velocity decreases, resulting in a lower kinetic energy of \( 498 \text{ J} \).

This decrease in kinetic energy is due to an inelastic interaction where non-conservative forces, like friction during the landing, cause a loss of energy in the form of heat or sound. Hence, while angular momentum is conserved, mechanical energy isn't in this scenario. Understanding these energy shifts is vital for analyzing situations where energy transformations occur during rotational motion.