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A man stands on a platform that is rotating (without friction) with an angular speed of \(1.2 \mathrm{rev} / \mathrm{s}\); his arms are outstretched and he holds a brick in each hand. The rotational inertia of the system consisting of the man, bricks, and platform about the central vertical axis of the platform is \(6.0 \mathrm{~kg} \cdot \mathrm{m}^{2}\). If by moving the bricks the man decreases the rotational inertia of the system to \(2.0 \mathrm{~kg} \cdot \mathrm{m}^{2}\), what are (a) the resulting angular speed of the platform and (b) the ratio of the new kinetic energy of the system to the original kinetic energy? (c) What source provided the added kinetic energy?

Short Answer

Expert verified
(a) 3.6 rev/s; (b) Energy ratio is 9:1; (c) Internal work done by the man pulling bricks.

Step by step solution

01

Understand the Concept of Conservation of Angular Momentum

The principle of conservation of angular momentum states that if no external torque acts on a system, its angular momentum is conserved. The formula is \( L = I \cdot \omega \), where \( L \) is angular momentum, \( I \) is the moment of inertia, and \( \omega \) is the angular speed.
02

Apply Conservation of Angular Momentum

Initially, the angular momentum \( L_i = I_i \cdot \omega_i = 6.0 \, \text{kg} \cdot \text{m}^2 \times 1.2 \, \text{rev/s} \). When the man changes the rotational inertia, the final angular momentum \( L_f = I_f \cdot \omega_f \) must equal \( L_i \). Since \( I_f = 2.0 \, \text{kg} \cdot \text{m}^2 \), set \( L_i = L_f \) and solve for \( \omega_f \).
03

Calculate the Resulting Angular Speed

Plug in the values: \( 6.0 \, \text{kg} \cdot \text{m}^2 \times 1.2 \, \text{rev/s} = 2.0 \, \text{kg} \cdot \text{m}^2 \times \omega_f \). Therefore, \( \omega_f = \frac{6.0 \times 1.2}{2.0} \), which gives \( \omega_f = 3.6 \, \text{rev/s} \).
04

Calculate the Original and New Kinetic Energy

The rotational kinetic energy is given by \( KE = \frac{1}{2} I \omega^2 \). Calculating the initial kinetic energy: \( KE_i = \frac{1}{2} \times 6.0 \, \text{kg} \cdot \text{m}^2 \times (2\pi\times1.2)^2 \). For the final kinetic energy, \( KE_f = \frac{1}{2} \times 2.0 \, \text{kg} \cdot \text{m}^2 \times (2\pi\times3.6)^2 \).
05

Calculate the Energy Ratio and Explanation

First, calculate \( KE_i \) and \( KE_f \), then find the ratio \( \frac{KE_f}{KE_i} \). Realize that the extra energy comes from the internal work done by the man when pulling the bricks towards the center.

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Key Concepts

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

Rotational Inertia
Rotational inertia, also known as the moment of inertia, is a key concept when dealing with rotating objects. It is essentially the rotational equivalent of mass in linear motion. Just like mass resists changes in velocity, rotational inertia resists changes in angular speed. This property depends on how the mass of an object is distributed in relation to its axis of rotation.

For example, the farther the mass is from the central axis, the higher the rotational inertia. In the exercise we’re looking at, the man initially has his arms outstretched with bricks. This leads to a rotational inertia of 6.0 kg·m². However, when he pulls the bricks inwards, closer to his body, the rotational inertia decreases to 2.0 kg·m². By understanding how to manipulate rotational inertia, you can change how an object behaves while rotating.

In practical terms, athletes like figure skaters use this principle to spin faster by pulling their limbs closer to their body, effectively reducing their rotational inertia and thus spinning at higher speeds without additional force.
Angular Speed
Angular speed denotes how fast an object is rotating around a particular axis. We often measure it in revolutions per second (rev/s) or radians per second (rad/s). The principle of angular momentum conservation lets us mathematically relate rotational inertia and angular speed.

In the provided exercise, the man's angular momentum is conserved. Initially, he spins at 1.2 rev/s with a rotational inertia of 6.0 kg·m². By changing this inertia to 2.0 kg·m², we apply the conservation of angular momentum: the product of the initial rotational inertia and angular speed must equal the product of the new rotational inertia and the new angular speed.

Using this principle, we calculate the resulting angular speed to be 3.6 rev/s after the change. So, even without any external force, the man's angular speed increases due to the manipulation of his body's rotational inertia.
Rotational Kinetic Energy
Rotational kinetic energy is the energy an object possesses due to its rotation, closely related to its mass distribution (rotational inertia) and its angular speed. The formula for rotational kinetic energy is given by \[ KE = \frac{1}{2} I \omega^2 \] where \( I \) is the rotational inertia, and \( \omega \) is the angular speed.

In our exercise, we first calculate the initial kinetic energy using the initial rotational inertia and angular speed: \[ KE_i = \frac{1}{2} \times 6.0 \, \text{kg} \cdot \text{m}^2 \times (2\pi \times 1.2)^2 \]
Then, we find the final kinetic energy with the reduced rotational inertia and increased angular speed:\[ KE_f = \frac{1}{2} \times 2.0 \, \text{kg} \cdot \text{m}^2 \times (2\pi \times 3.6)^2 \]
Calculating these, we can determine the ratio of the new kinetic energy to the original kinetic energy by dividing \( KE_f \) by \( KE_i \). The interesting revelation here is that the extra energy comes from the internal work done by the man—illustrating the basic principles of physics where energy transformation within a closed system can result in increased motion.

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Most popular questions from this chapter

Force \(\vec{F}=(2.0 \mathrm{~N}) \hat{\mathrm{i}}-(3.0 \mathrm{~N}) \hat{\mathrm{k}}\) acts on a pebble with position vector \(\vec{r}=(0.50 \mathrm{~m}) \mathrm{j}-(2.0 \mathrm{~m}) \mathrm{k}\) relative to the origin. In unitvector notation, what is the resulting torque on the pebble about (a) the origin and (b) the point \((2.0 \mathrm{~m}, 0,-3.0 \mathrm{~m}) ?\)

A solid ball of radius \(6.0 \mathrm{~cm}\) is initially rolling smoothly at 10 \(\mathrm{m} / \mathrm{s}\) along a horizontal floor. It then rolls smoothly up a ramp until it momentarily stops. What maximum height above the floor does it reach?

The angular momentum of a flywheel having a rotational inertia of \(0.140 \mathrm{~kg} \cdot \mathrm{m}^{2}\) about its central axis decreases from \(3.00\) to \(0.800 \mathrm{~kg} \cdot \mathrm{m}^{2} / \mathrm{s}\) in \(1.50 \mathrm{~s}\). (a) What is the magnitude of the average torque acting on the flywheel about its central axis during this period? (b) Assuming a constant angular acceleration, through what angle does the flywheel turn? (c) How much work is done on the wheel? (d) What is the average power of the flywheel?

Two disks are mounted (like a merry-go-round) on lowfriction bearings on the same axle and can be brought together so that they couple and rotate as one unit. The first disk, with rotational inertia \(3.30 \mathrm{~kg} \cdot \mathrm{m}^{2}\) about its central axis, is set spinning counterclockwise at \(450 \mathrm{rev} / \mathrm{min}\). The second disk, with rotational inertia \(6.60 \mathrm{~kg} \cdot \mathrm{m}^{2}\) about its central axis, is set spinning counterclockwise at \(900 \mathrm{rev} / \mathrm{min}\). They then couple together. (a) What is their angular speed after coupling? If instead the second disk is set spinning clockwise at \(900 \mathrm{rev} / \mathrm{min}\), what are their (b) angular speed and (c) direction of rotation after they couple together?

A Texas cockroach of mass \(0.20 \mathrm{~kg}\) runs counterclockwise around the rim of a lazy Susan (a circular disk mounted on a vertical axle) that has radius \(18 \mathrm{~cm}\), rotational inertia \(5.0 \times 10^{-3} \mathrm{~kg} \cdot \mathrm{m}^{2}\), and frictionless bearings. The cockroach's speed (relative to the ground) is \(2.0 \mathrm{~m} / \mathrm{s}\), and the lazy Susan turns clockwise with angular speed \(\omega_{0}=2.8 \mathrm{rad} / \mathrm{s}\). The cockroach finds a bread crumb on the rim and, of course, stops. (a) What is the angular speed of the lazy Susan after the cockroach stops? (b) Is mechanical energy conserved as it stops?

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