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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 \(70.0-\mathrm{kg}\) 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?

Short Answer

Expert verified
(a) The angular speed after the parachutist lands is approximately 1.39 rad/s. (b) Before: 1080 J, After: 501 J; energy is lost due to inelastic collision.

Step by step solution

01

Understand the System Before the Parachutist Lands

Initially, the turntable is the only object in motion. It's a rotating disk with a radius \( R = 2.00 \, \mathrm{m} \) and mass \( M = 120 \, \mathrm{kg} \). Its initial angular speed is \( \omega_i = 3.00 \, \mathrm{rad/s} \). We calculate its moment of inertia \( I_i = \frac{1}{2}MR^2 \).
02

Calculate Initial Moment of Inertia of the Turntable

The moment of inertia for a uniform disk rotating about its center is given by \( I_i = \frac{1}{2}MR^2 \). Substitute \( M = 120 \, \mathrm{kg} \) and \( R = 2.00 \, \mathrm{m} \) to get:\[ I_i = \frac{1}{2} \times 120 \, \mathrm{kg} \times (2.00 \, \mathrm{m})^2 = 240 \, \mathrm{kg \cdot m^2} \]
03

Compute the Initial Kinetic Energy of the Turntable

The rotational kinetic energy before the parachutist lands is given by:\[ KE_i = \frac{1}{2} I_i \omega_i^2 = \frac{1}{2} \times 240 \, \mathrm{kg \cdot m^2} \times (3.00 \, \mathrm{rad/s})^2 = 1080 \, \mathrm{J} \]
04

Understand the System After the Parachutist Lands

After landing, the parachutist adds to the system mass and affects its moment of inertia. Treat the parachutist as a particle located at the radius \( R = 2.00 \, \mathrm{m} \). The system's new moment of inertia \( I_f = I_i + m R^2 \), where \( m = 70.0 \, \mathrm{kg} \) is the parachutist's mass.
05

Calculate New Moment of Inertia Including the Parachutist

Using \( I_f = I_i + m R^2 \), compute:\[ I_f = 240 \, \mathrm{kg \cdot m^2} + 70.0 \, \mathrm{kg} \times (2.00 \, \mathrm{m})^2 = 520 \, \mathrm{kg \cdot m^2} \]
06

Apply Conservation of Angular Momentum to Find Final Angular Speed

Initially, the angular momentum \( L_i = I_i \omega_i \). When the parachutist lands, angular momentum is conserved:\[ I_i \omega_i = I_f \omega_f \]Substitute known values:\[ 240 \, \mathrm{kg \cdot m^2} \times 3.00 \, \mathrm{rad/s} = 520 \, \mathrm{kg \cdot m^2} \times \omega_f \]Solve for \( \omega_f \):\[ \omega_f = \frac{720}{520} \, \mathrm{rad/s} = 1.385 \, \mathrm{rad/s} \approx 1.39 \, \mathrm{rad/s} \]
07

Compute the Final Kinetic Energy After Parachutist Lands

The kinetic energy after landing is calculated using the new angular speed:\[ KE_f = \frac{1}{2} I_f \omega_f^2 = \frac{1}{2} \times 520 \, \mathrm{kg \cdot m^2} \times (1.39 \, \mathrm{rad/s})^2 = 500.99 \, \mathrm{J} \approx 501 \, \mathrm{J} \]
08

Analyze Why Kinetic Energies Are Not Equal

The kinetic energy reduces because some kinetic energy is converted to other forms of energy, like heat or sound, during the collision when the parachutist lands. This process is inelastic, conserving angular momentum but not mechanical energy.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Moment of Inertia
The moment of inertia is a fundamental concept in rotational dynamics, playing a role similar to mass in linear motion. It measures an object's resistance to changes in its rotation. The moment of inertia depends on the mass distribution relative to the axis of rotation.
For a uniform disk rotating about its center, the moment of inertia is calculated as \( I = \frac{1}{2} M R^2 \). Here, \( M \) is the mass of the disk and \( R \) its radius. In our example, the disk has a mass of 120 kg and a radius of 2 m, hence the initial moment of inertia is 240 kg·m².
The presence of additional mass, such as a parachutist landing on the disk, changes the moment of inertia. This change is computed by adding the inertia of the parachutist (treated as a point mass at the radius): \( mR^2 \). Hence, the final moment of inertia of the system becomes 520 kg·m², taking the added mass into account.
Rotational Kinetic Energy
Rotational kinetic energy is the energy an object possesses due to its rotation. It is defined similarly to linear kinetic energy but uses the rotational quantities:
  • The formula for rotational kinetic energy is \( KE = \frac{1}{2} I \omega^2 \), where \( I \) is the moment of inertia and \( \omega \) the angular velocity.
The kinetic energy of our disk before the parachutist lands is 1080 J, derived using its initial moment of inertia and angular speed of 3 rad/s.
After the parachutist lands, the system's kinetic energy decreases to about 501 J. This decrease occurs because changes in the system, such as an increase in moment of inertia and a decrease in angular velocity, lead to an adjustment in kinetic energy. The previously greater kinetic energy is not entirely conserved.
Inelastic Collision
An inelastic collision is a type of collision where kinetic energy is not conserved. This happens in our exercise when the parachutist lands softly on the rotating turntable.
During an inelastic collision, some of the kinetic energy is transformed into other forms of energy, such as thermal energy or sound. Despite this loss, the principle of angular momentum conservation still applies. It ensures that the total angular momentum before and after the collision remains the same.
In our scenario, while the angular speed drops following the parachutist's landing, the overall angular momentum remains unchanged. Thus, we observe a reduction in kinetic energy from 1080 J to approximately 501 J due to the energy transformation inherent in inelastic collisions.
Uniform Disk
A uniform disk refers to a flat, circular object with mass distributed evenly across its entire surface. This distribution leads to specific properties that affect rotational motion.
In our exercise, the turntable is a uniform disk with a radius of 2 meters and a total mass of 120 kg. This uniformity allows us to apply standard formulas for calculating moment of inertia, which for a disk is \( \frac{1}{2} M R^2 \).
  • The symmetry and consistent mass distribution of a uniform disk simplify these calculations and make it a frequently encountered object in physics problems.
  • Such simplifications become crucial when applying concepts of rotation and angular momentum, as seen in determining the effects of added mass, like our parachutist example.
This ensures that the calculations remain straightforward and manageable when learning about rotational dynamics.

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