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Chapter 12: Problem 31

The rotational 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 rotational 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?

Short Answer

Expert verified
a) 1.47 N路m. b) -17.69 rad. c) 29.85 J. d) 19.9 W.

Step by step solution

01

Identify initial and final angular momentum

The initial angular momentum, \(L_i\), is given as \(3.00\, \mathrm{kg \cdot m^2 / s}\) and the final angular momentum, \(L_f\), is given as \(0.80\, \mathrm{kg \cdot m^2 / s}\).
02

Calculate change in angular momentum

Change in angular momentum, \(\Delta L\), can be determined by subtracting the final angular momentum from the initial angular momentum: \[ \Delta L = L_f - L_i \] \[ \Delta L = 0.80 \ \text{kg} \cdot \text{m}^2/\text{s} - 3.00 \ \text{kg} \cdot \text{m}^2/\text{s} \] \[ \Delta L = -2.20 \ \text{kg} \cdot \text{m}^2/\text{s} \]
03

Determine the average torque

The average torque, \(\tau_{avg}\), can be computed using the formula: \[ \tau_{avg} = \frac{\Delta L}{\Delta t} \] Given \(\Delta t = 1.50 \ \text{s}\), \[ \tau_{avg} = \frac{-2.20}{1.50} \] \[ \tau_{avg} = -1.47 \ \text{N} \cdot \text{m} \] Its magnitude is \(1.47 \ \text{N} \cdot \text{m} \).
04

Calculate angular acceleration

Using the rotational inertia, \(I = 0.140 \ \text{kg} \cdot \text{m}^2\), and \(\Delta \omega = \frac{\Delta L}{I}\), \[ \Delta \omega = \frac{-2.20 \ \text{kg} \cdot \text{m}^2 /\text{s}}{0.140 \ \text{kg} \cdot \text{m}^2} \] \[ \Delta \omega = -15.71 \ \text{rad}/\text{s}^2 \]
05

Calculate the angle turned

Using the equation \(\theta = \frac{\Delta \omega \cdot t^2}{2}\), \[ \theta = \frac{(-15.71) \cdot (1.50)^2}{2} \] \[ \theta = -17.69 \ \text{rad} \]
06

Calculate the work done

Work done, \(W\), is given by the change in rotational kinetic energy, \(\Delta K = \frac{L_i^2}{2I} - \frac{L_f^2}{2I}\), \[ W = \frac{(3.00)^2}{2 \cdot 0.14} - \frac{(0.80)^2}{2 \cdot 0.14} \] \[ W = 32.14 - 2.29 \] \[ W = 29.85 \ \text{J} \]
07

Calculate average power

Average power, \(P_{avg}\), is given by: \[ P_{avg} = \frac{W}{\Delta t} \] \[ P_{avg} = \frac{29.85}{1.50} \] \[ P_{avg} = 19.9 \ \text{W} \]

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Key Concepts

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

Angular Momentum
Angular momentum is a measure of the rotational motion of an object. It's similar to linear momentum but for rotating systems. The initial angular momentum (\(L_i\)) and final angular momentum (\(L_f\)) provide insight into how an object's rotational state changes over time. In this exercise, angular momentum decreases from 3.00 to 0.80 \(kg \bullet m^2 / s\). This change is crucial to understanding how other aspects, like torque and work, will be affected.

It's given by: \[L = I \bullet \text{蠅} \] where \(I\) is the rotational inertia and \(蠅\) is the angular velocity.
Torque
Torque is a measure of the turning force on an object. It's the rotational equivalent of force. The average torque (\(\tau_{avg}\)) acting on the flywheel can be calculated using the change in angular momentum (\(\Delta L\)) over time (\(\Delta t\)).

The formula is:
\[\tau_{avg} = \frac{\Delta L}{\Delta t}\]

When we plug in the values:
\[\tau_{avg} = \frac{-2.20 \, kg \bullet m^2 / s}{1.50 \, s} = -1.47 \, N \bullet m\]

It's important to note that while the torque is negative, the magnitude is always a positive value. This shows us the force used to reduce the rotational momentum in this scenario.
Angular Acceleration
Angular acceleration is the rate of change in angular velocity. In this context, it helps us understand how quickly the flywheel's rotational speed is changing. Using the rotational inertia (\(I\, = \ 0.140 \, kg \bullet m^2\) and the change in angular velocity (\( \Delta \text{蠅}\)), we find the angular acceleration.

The formula used is:
\[\Delta \text{蠅} = \frac{\Delta L}{I}\]

Plugging in our values, we get:
\[\Delta \text{蠅} = \frac{-2.20 \, kg \bullet m^2 / s}{0.140 \, kg \bullet m^2} = -15.71 \, rad/s^2\]

This negative value indicates the direction of the angular acceleration opposite to the initial motion.
Rotational Inertia
Rotational inertia, also known as the moment of inertia, measures how difficult it is to change the rotational motion of an object. It's the rotational equivalent of mass in linear motion. For the flywheel, the rotational inertia (\(I\)) is given as 0.140 \ kg \bullet m^2. This value shows the resistance of the flywheel to changes in its rotational motion.

Higher rotational inertia means more torque is required to accelerate or decelerate the object鈥檚 rotation.

Rotational inertia is calculated differently based on the object's shape and axis of rotation.
Work Done
Work done in rotational motion is related to the change in rotational kinetic energy. Here, the work (\(W\)) on the flywheel can be calculated by the difference in kinetic energy from start to end.
The formula used is:
\[ W = \frac{L_i^2}{2I} - \frac{L_f^2}{2I} \]
Plugging in the values:
\[W = \frac{(3.00 \, kg \bullet m^2 / s)^2}{2 \bullet 0.140 \, kg \bullet m^2} - \frac{(0.80 \, kg \bullet m^2 / s)^2}{2 \bullet 0.140 \, kg \bullet m^2}\]
\[W = 32.14 - 2.29 = 29.85 \, J\] While work done in rotational motion could be complex, this approach simplifies calculations by focusing on energy change.
Power
Power in rotational motion refers to the rate at which work is done. The average power (\(P_{avg}\)) of the flywheel can be calculated by dividing the total work done by the time taken.
The formula is:
\[ P_{avg} = \frac{W}{\Delta t}\] Plugging in the values:
\[ P_{avg} = \frac{29.85 \, J}{1.50 \, s} = 19.9 \, W\]
Understanding power helps to gauge the efficiency and performance of the system in transforming energy. This information is often applied in engineering and machinery design to improve performance.

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Most popular questions from this chapter

A \(2.0 \mathrm{~kg}\) particle-like object moves in a plane with velocity components \(v_{x}=30 \mathrm{~m} / \mathrm{s}\) and \(v_{y}=60 \mathrm{~m} / \mathrm{s}\) as it passes through the point with \((x, y)\) coordinates of \((3.0,-4.0) \mathrm{m}\). Just then, what is its rotational momentum relative to (a) the origin and (b) the point \((-2.0,-2.0) \mathrm{m}\) ?

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?

The rotational inertia of a collapsing spinning star changes to \(\frac{1}{3}\) its initial value. What is the ratio of the new rotational kinetic energy to the initial rotational kinetic energy?

A hollow sphere of radius \(0.15 \mathrm{~m}\), with rotational inertia \(I=0.040 \mathrm{~kg} \cdot \mathrm{m}^{2}\) about a line through its center of mass, rolls without slipping up a surface inclined at \(30^{\circ}\) to the horizontal. At a certain initial position, the sphere's total kinetic energy is \(20 \mathrm{~J}\). (a) How much of this initial kinetic energy is rotational? (b) What is the speed of the center of mass of the sphere at the initial position? What are (c) the total kinetic energy of the sphere and (d)) the speed of its center of mass after it has moved \(1.0 \mathrm{~m}\) up along the incline from its initial position?

What torque about the origin acts on a particle moving in the \(x y\) plane, clockwise about the origin, if the particle has the following magnitudes of rotational momentum about the origin: (a) \(4.0 \mathrm{~kg} \cdot \mathrm{m}^{2} / \mathrm{s}\) (b) \(\left(4.0 \frac{1}{s}\right) t^{2} \mathrm{~kg} \cdot \mathrm{m}^{2} / \mathrm{s}\) (c) \(\left(4.0 \frac{1}{s^{1}}\right) \sqrt{t} \mathrm{~kg} \cdot \mathrm{m}^{2} / \mathrm{s}\) (d) \(\left(4.0 \mathrm{~s}^{2}\right) / t^{2} \mathrm{~kg} \cdot \mathrm{m}^{2} / \mathrm{s}\) ?
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