A turntable of radius \(25 \mathrm{cm}\) and rotational inertia \(0.0154 \mathrm{kg} \cdot \mathrm{m}^{2}\) is spinning freely at 22.0 rpm about its central axis, with a \(19.5-\mathrm{g}\) mouse on its outer edge. The mouse walks from the edge to the center. Find (a) the new rotation speed and (b) the work done by the mouse.

Short Answer

Expert verified
The new rotation speed \(\omega_2\) is found after substituting the given and calculated values into \(\omega_2 = I_1 * \omega_1 / I_2 \), and the result for the work done by the mouse \(W\) can be gotten by substituting the initial and final kinetic energy into \(W = KE_2 - KE_1\).

Step by step solution

01

Physiology of Moment of Inertia and Angular Speed

The moment of inertia of an object rotating about an axis can vary depending on the distribution of mass about the axis of rotation. Thus, initially, the rotation inertia \(I_1 = 0.0154 \, kg \cdot m^{2}\) includes the inertias of both the turntable and the mouse located at its edge, and the angular speed \(\omega_1\) is provided in rpm, which can be translated into rad/s as \(\omega_1 = 2\pi*(22.0/60) = 2.309 \, rad/s.\) The initial angular momentum \(L_1\) can be given as \(I_1 * \omega_1\).
02

Calculate the Moment of Inertia with the Mouse at Center

After the mouse walks towards the center, the moment of inertia \(I_2\) of the system decreases because the mouse's mass is no longer located far from the rotational axis. The new moment of inertia \(I_2\) is the inertia of the turntable alone since the mouse is now at the rotational axis and contributes no inertia. Using conservation of angular momentum, we find the new angular speed as \(I_1 * \omega_1 = I_2 * \omega_2 => \omega_2 = I_1 * \omega_1 / I_2 \).
03

Calculate the Work Done by the Mouse

The work done by the mouse can be found by comparing the initial and final kinetic energy of the system and computing the difference between them. The initial rotational kinetic energy of the system is \( KE_1 = 0.5 * I_1 * \omega_1^{2}\) and the final kinetic energy \( KE_2 = 0.5 * I_2 * \omega_2^{2}\). The work done is then \(W = KE_2 - KE_1\).

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Key Concepts

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

Moment of Inertia
When we discuss the dynamics of rotational motion, the moment of inertia is a key concept, akin to the role of mass in linear motion. It quantifies an object's resistance to changes in its rotational speed, and knowing this resistance value is crucial for solving many physics problems involving rotation.

Moment of inertia is heavily influenced by two factors: the mass of the objects and how this mass is distributed relative to the axis of rotation. It's described mathematically as the sum of the products of the mass elements and the square of their distance from the rotation axis: \( I = \sum m_i r_i^2 \). In our textbook example, the turntable and mouse system initially has a larger moment of inertia because the mouse, a mass element, is located at the edge, at a distance from the axis. Once the mouse moves to the center, the overall moment of inertia decreases, as this redistribution of mass decreases the resistance to rotational acceleration.

Demonstrating the Dependency

The exercise improvement advice here is to visualize this concept. For instance, if you hold a spinning bicycle wheel with its axle horizontal and then tilt the axle, you will feel resistance. This is because of the moment of inertia, which depends on both the mass and the distribution of the mass.
Angular Momentum Conservation
Angular momentum conservation is a principle that helps us predict the behavior of rotating systems. It's vital in situations where no external torques are acting on a system, meaning the total angular momentum remains constant over time. This principle can be stated as \( L_1 = L_2 \) or \( I_1 \omega_1 = I_2 \omega_2 \), where \(I_1\) and \(\omega_1\) are the initial moment of inertia and angular velocity, respectively, and \(I_2\) and \(\omega_2\) are the final values.

In our turntable example, as the mouse moves towards the center, there are no external torques, thereby the angular momentum of the system is conserved. Initially, the system had a higher moment of inertia and a lower angular velocity. As the mouse reaches the center, the moment of inertia decreases, and to conserve angular momentum, the angular velocity increases.

Real-World Application

Ice skaters use this principle to control their spin. When a skater pulls in their arms, they reduce their moment of inertia and spin faster to conserve angular momentum. This illustrates the inverse relationship between the moment of inertia and angular velocity in the absence of external torques.
Rotational Kinetic Energy
Just as objects in linear motion have kinetic energy due to their mass and velocity (\( KE = \frac{1}{2}mv^2 \)), objects in rotational motion have rotational kinetic energy because of their moment of inertia and angular velocity. The formula for rotational kinetic energy is \( KE = \frac{1}{2}I\omega^2 \), where \(I\) is the moment of inertia and \(\omega\) is the angular velocity.

In the turntable scenario, the change in the mouse's position does not only alter the system's angular velocity due to conservation of angular momentum; it also changes the system's rotational kinetic energy. As the turntable's rotation speeds up, its rotational kinetic energy increases, even though no external forces or energies are added to the system. The work done by the mouse, which involves moving against rotational inertia, is what changes the energy of the system. Therefore, the mouse is essentially converting its own energy to increase the turntable's rotational kinetic energy.

Understanding Energy Transfer

For deeper comprehension, consider a diver tucking into a spin during a dive. Their tucked position minimizes the moment of inertia, increasing angular velocity, and thus increasing rotational kinetic energy as they spin towards the water. This conservation and transformation of energy are at the heart of rotational dynamics and critical to understanding motion in a rotating frame.

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Most popular questions from this chapter

A baseball player extends his arm straight up to catch a \(145-\mathrm{g}\) baseball moving horizontally at \(42 \mathrm{m} / \mathrm{s}\). It's \(63 \mathrm{cm}\) from the player's shoulder joint to the point the ball strikes his hand, and his arm remains stiff while it rotates about the shoulder during the catch. The player's hand recoils \(5.00 \mathrm{cm}\) horizontally while he stops the ball. What average torque does the player's arm exert on the ball?

A time-dependent torque given by \(\tau=a+b\) sin \(c t\) is applied to an object that's initially stationary but is free to rotate. Here \(a, b\) and \(c\) are constants. Find an expression for the object's angular momentum as a function of time, assuming the torque is first applied at \(t=0\).

Express the units of angular momentum (a) using only the fundamental units kilogram, meter, and second; (b) in a form involving newtons; (c) in a form involving joules.

Tornadoes in the northern hemisphere rotate counterclockwise as viewed from above. A far-fetched idea suggests that driving on the right side of the road may increase the frequency of tornadoes. Does this idea have any merit? Explain in terms of the angular momentum imparted to the air as two cars pass.

A skater has rotational inertia \(4.2 \mathrm{kg} \cdot \mathrm{m}^{2}\) with his fists held to his chest and \(5.7 \mathrm{kg} \cdot \mathrm{m}^{2}\) with his arms outstretched. The skater is spinning at 3.0 rev/s while holding a 2.5 -kg weight in each outstretched hand; the weights are \(76 \mathrm{cm}\) from his rotation axis. If he pulls his hands in to his chest, so they're essentially on his rotation axis, how fast will he be spinning?

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