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Chapter 12: Problem 37

A man stands on a platform that is rotating (without friction) with a rotational 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 axis is \(6.0 \mathrm{~kg} \cdot \mathrm{m}^{2} .\) If by moving the bricks the man decreases the rotational inertia of the system to \(2.0 \mathrm{~kg} \cdot \mathrm{m}^{2}\), (a) what is the resulting rotational speed of the platform and (b) what is the ratio of the new kinetic energy of the system to the original kinetic energy? (c) What provided the added kinetic energy?

Short Answer

Expert verified
The resulting rotational speed is 3.6 rev/s. The ratio of new kinetic energy to the initial kinetic energy is 3. The added kinetic energy was provided by the work done by the man in moving the bricks.

Step by step solution

01

- Understand the problem

We need to determine the new rotational speed when the rotational inertia changes and the ratio of the new kinetic energy to the original kinetic energy. Then, we need to identify the source of the added kinetic energy.
02

- Use conservation of angular momentum

The system's angular momentum is conserved because there is no external torque. The initial angular momentum is \[ L_i = I_i \times \text{蠅}_i \] where \( I_i = 6.0 \, \text{kg} \cdot \text{m}^2 \) and \( \text{蠅}_i = 1.2 \, \text{rev/s} \).
03

- Calculate initial angular momentum

Convert the initial angular speed to rad/s: \( \text{蠅}_i = 1.2 \, \text{rev/s} \times 2\text{蟺} \, \text{rad/rev} = 2.4\text{蟺} \, \text{rad/s} \)Then: \[ L_i = 6.0 \, \text{kg} \cdot \text{m}^2 \times 2.4\text{蟺} \, \text{rad/s} = 14.4\text{蟺} \, \text{kg} \cdot \text{m}^2/s \]
04

- Apply conservation of angular momentum to find resulting angular speed

The final angular momentum \(L_f\) must equal the initial angular momentum: \( L_f = I_f \times \text{蠅}_f \)where \( I_f = 2.0 \, \text{kg} \cdot \text{m}^2 \). Since \( L_f = L_i \), we have: \[ I_f \times \text{蠅}_f = L_i \]\[ 2.0 \, \text{kg} \cdot \text{m}^2 \times \text{蠅}_f = 14.4\text{蟺} \, \text{kg} \cdot \text{m}^2/s \]Solve for \( \text{蠅}_f \): \[ \text{蠅}_f = \frac{14.4\text{蟺}}{2.0} = 7.2\text{蟺} \, \text{rad/s} \]
05

- Convert resulting angular speed to rev/s

Convert \( \text{蠅}_f \) from rad/s to rev/s: \[ \text{蠅}_f = \frac{7.2\text{蟺} \, \text{rad/s}}{2\text{蟺} \, \text{rad/rev}} = 3.6 \, \text{rev/s} \]
06

- Calculate initial kinetic energy

The initial rotational kinetic energy is given by \[ K_i = \frac{1}{2} I_i \text{蠅}_i^2 \]Substitute \( I_i = 6.0 \, \text{kg} \cdot \text{m}^2 \) and \( \text{蠅}_i = 2.4\text{蟺} \, \text{rad/s} \): \[ K_i = \frac{1}{2} \times 6.0 \times (2.4 \text{蟺})^2 = 54.4 \, \text{J} \]
07

- Calculate final kinetic energy

The final rotational kinetic energy is given by \[ K_f = \frac{1}{2} I_f \text{蠅}_f^2 \]Substitute \( I_f = 2.0 \, \text{kg} \cdot \text{m}^2 \) and \( \text{蠅}_f = 7.2\text{蟺} \, \text{rad/s} \): \[ K_f = \frac{1}{2} \times 2.0 \times (7.2\text{蟺})^2 = 163.2 \, \text{J} \]
08

- Calculate the ratio of new kinetic energy to initial kinetic energy

The ratio of the new kinetic energy to the initial kinetic energy is given by \[ \frac{K_f}{K_i} = \frac{163.2}{54.4} \approx 3.0 \]
09

- Identify the source of added kinetic energy

The man provided the added kinetic energy by doing work as he moved the bricks closer to the central axis, reducing the system's rotational inertia.

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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 called moment of inertia, is a property of an object that quantifies how much it resists rotational motion.
This means it's harder to start or stop an object's rotation if it has a high rotational inertia.
Mathematically, rotational inertia (\text{I}) is the sum of each mass multiplied by the square of its distance from the axis of rotation:
\[{I} = \sum{m_i r_i^2}\] where m is the mass and r is the distance from the axis.
The farther the mass is from the rotational axis, the higher the rotational inertia.
In the exercise, the system initially has a rotational inertia of 6.0 \text{kg}路m虏, which decreases to 2.0 \text{kg}路m虏 when the man pulls the bricks closer.
By reducing the distance of the bricks from the axis, the rotational inertia decreases significantly.
Rotational Speed
Rotational speed is how quickly an object rotates or spins around an axis.
It's specified in units like revolutions per second (rev/s) or radians per second (rad/s).
The relationship between the two units is: 1 revolution = 2\text{蟺} radians.
In the given problem, the man's initial rotational speed is 1.2 rev/s.
This speed increases to 3.6 rev/s when the rotational inertia decreases.
According to the conservation of angular momentum, the product of rotational inertia (\text{I}) and rotational speed (\text{蠅}) remains constant if no external torque is applied:
\[{L} = {I} {蠅} = \text{constant} \]\ This principle means when \text{I} decreases, \text{蠅} must increase to maintain the same angular momentum.
Kinetic Energy
Kinetic energy in a rotating system is known as rotational kinetic energy.
It's given by the formula: \[{K} = \frac{1}{2} {I} {蠅^2}\] where \text{K} is the rotational kinetic energy, \text{I} is the rotational inertia, and \text{蠅} is the rotational speed.
In the exercise, the initial kinetic energy is: \[{K_i} = \frac{1}{2} \times 6.0 \times (2.4\text{蟺})^2 \approx 54.4 \text{J}\],
When the rotational inertia decreases to 2.0 \text{kg}\text{m}^{2}, the kinetic energy becomes: \[{K_f} = \frac{1}{2} \times 2.0 \times (7.2\text{蟺})^2\approx 163.2\]\ The ratio of the final to initial kinetic energy is approximately 3.0.
The man provides the extra kinetic energy by doing work as he pulls the bricks inward, causing his body to spin faster.

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Most popular questions from this chapter

At time \(t=0\), a \(2.0 \mathrm{~kg}\) particle has position vector \(\vec{r}=(4.0 \mathrm{~m}) \hat{\mathrm{i}}-(2.0 \mathrm{~m}) \hat{\mathrm{j}}\) relative to the origin. Its velocity just then is given by \(\vec{v}=\left(-6.0 \mathrm{~m} / \mathrm{s}^{3}\right) t^{2} \hat{\mathrm{i}}\). About the origin and for \(t>0\), what are (a) the particle's rotational momentum and (b) the torque acting on the particle? (c) Repeat (a) and (b) about a point with coordinates \((-2.0,-3.0,0.0) \mathrm{m}\) instead of about the origin.

Two particles, each of mass \(m\) and speed \(v\), travel in opposite directions along parallel lines separated by a distance \(d\). (a) In terms of \(m, v\), and \(d\), find an expression for the magnitude \(L\) of the rotational momentum of the two-particle system around a point midway between the two lines. (b) Does the expression change if the point about which \(L\) is calculated is not midway between the lines? (c) Now reverse the direction of travel for one of the particles and repeat (a) and (b).

A horizontal platform in the shape of a circular disk rotates on a frictionless bearing about a vertical axle through the center of the disk. The platform has a mass of \(150 \mathrm{~kg}\), a radius of \(2.0 \mathrm{~m}\), and a rotational inertia of \(300 \mathrm{~kg} \cdot \mathrm{m}^{2}\) about the axis of rotation. A \(60 \mathrm{~kg}\) student walks slowly from the rim of the platform toward the center. If the rotational speed of the system is \(1.5 \mathrm{rad} / \mathrm{s}\) when the student starts at the rim, what is the rotational speed when she is \(0.50 \mathrm{~m}\) from the center?

Consider a \(66-\mathrm{cm}\) -diameter tire on a car traveling at 80 \(\mathrm{km} / \mathrm{h}\) on a level road in the positive direction of an \(x\) axis. Relative to a woman in the car, what are (a) the translational velocity \(\vec{v}_{\text {center }}\) and (b) the magnitude \(a_{\text {center }}\) of the translational acceleration of the center of the wheel? What are (c) \(\vec{v}_{\text {top }}\) and (d) \(a_{\text {top }}\) for a point at the top of the tire? What are (e) \(\vec{v}_{\text {bot }}\) and (f) \(a_{\text {bot }}\) for a point at the bottom of the tire? Now repeat the questions relative to a hitchhiker sitting near the road: What are \((\mathrm{g}) \vec{v}\) at the wheel's center, \((\mathrm{h}) a\) at the wheel's center, (i) \(\vec{v}\) at the tire top, (j) \(a\) at the tire top, \((\mathrm{k}) \vec{v}\) at the tire bottom, and (1) \(a\) at the tire bottom?

56\. Uniform Rod In Fig. \(12-34\), a uniform rod (length \(=0.60 \mathrm{~m}\), mass \(1.0 \mathrm{~kg}\) ) rotates about an axis through one end, with a rotational inertia of \(0.12 \mathrm{~kg} \cdot \mathrm{m}^{2}\). As the rod swings through its lowest position, the end of the rod collides with a small \(0.20\) \(\mathrm{kg}\) putty wad that sticks to the end of the rod. If the rotational speed of the rod just before the collision is \(2.4 \mathrm{rad} / \mathrm{s}\), what is the rotational speed of the rod-putty system immediately after the collision?
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