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A wheel rotates clockwise about its central axis with an angular momentum of \(600 \mathrm{~kg} \cdot \mathrm{m}^{2} / \mathrm{s}\). At time \(t=0\), a torque of magnitude \(50 \mathrm{~N} \cdot \mathrm{m}\) is applied to the wheel to reverse the rotation. At what time \(t\) is the angular speed zero?

Short Answer

Expert verified
The angular speed becomes zero at 12 seconds.

Step by step solution

01

Understand the Relation Between Torque and Angular Momentum

The change in angular momentum is determined by the torque applied. According to Newton's second law for rotation, the torque \( \tau \) applied to an object is related to the rate of change of its angular momentum \( L \) by the equation \( \tau = \frac{dL}{dt} \).
02

Calculate the Time to Change Angular Momentum to Zero

The problem states that we need the angular speed to become zero, which means the initial angular momentum must reach zero. The change in angular momentum (\( \Delta L \)) can also be expressed as \( \Delta L = \tau \cdot t \). The process will take the time period where \( \Delta L = -L_0 \), where \( L_0 \) is the initial angular momentum.
03

Substitute Known Values and Solve for Time

Given the initial angular momentum \( L_0 = 600 \mathrm{~kg} \cdot \mathrm{m}^2 / \mathrm{s} \) and the torque \( \tau = 50 \mathrm{~N} \cdot \mathrm{m} \), substitute these into the equation \( 50 \cdot t = -600 \). Solve for \( t \) by rearranging the equation to \( t = - \frac{600}{50} = 12 \).
04

Interpret the Result in Context of the Problem

Since we obtained a negative value during portion of calculations that represents time to stop and reverse direction, the final answer for time \( t \) is positive, representing the duration after \( t=0\). Thus, the wheel's angular speed reaches zero 12 seconds after the torque is applied.

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Key Concepts

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

Torque
Torque is an essential concept in understanding rotational motion. It acts like a twist that causes an object to rotate. The amount of torque exerted depends on two main factors:
  • The force applied.
  • The distance from the pivot point where the force is applied (also known as the moment arm).
The equation to express torque mathematically is \( \tau = r \cdot F \cdot \sin(\theta) \), where \( \tau \) is the torque,
\( r \) is the moment arm, \( F \) is the force applied, and \( \theta \) is the angle between the force and the moment arm.
Torque plays a critical role in changing an object's rotational motion. An applied torque can alter the angular momentum, causing it to increase or decrease, as seen in the wheel example. When torque is applied to the wheel, it changes the direction of the angular momentum, illustrating how torque can influence rotational motion. This understanding is crucial when solving problems involving rotational dynamics.
Angular Speed
Angular speed is a measure of how fast an object rotates or revolves around a specific point or axis. It's akin to the linear speed but for rotational motion. Angular speed can be described as the angle covered per unit time, and it is typically measured in radians per second (\( \text{rad/s} \)).
To grasp the concept further, imagine the hands of a clock. As the minute hand sweeps around to point back to the same number, it completes one full rotation. The rate at which it completes that rotation is its angular speed.
In the context of the wheel problem, the focus was to reverse the wheel's direction and reduce its angular speed to zero. This means the wheel's initial angular momentum must bring about a complete stop, allowing the torque to reverse the motion.
Understanding angular speed is vital. It helps in analyzing how quickly an object's angular position changes over time, emphasizing the dynamics of rotational systems.
Newton's Second Law for Rotation
Newton's second law for rotation provides a fundamental principle for understanding how rotational motion operates. Just as Newton's second law for linear movement describes how force affects acceleration, its rotational counterpart illustrates the relationship between torque and angular acceleration.
The law can be summarized by the formula: \( \tau = I \cdot \alpha \). Here, \( \tau \) represents torque, \( I \) stands for the moment of inertia (a measure of how much an object resists rotational acceleration), and \( \alpha \) is the angular acceleration.
In the wheel example, we see this law in action. The applied torque of 50 N·m is responsible for changing the wheel's angular momentum over time, indicating that the external force applied induces an angular acceleration.
This law is crucial for predicting how external forces will influence rotational motion. It helps explain why an object will start rotating faster or slower, depending on the torque applied and the object's inertia. It essentially bridges the gap between force, mass, and motion within the realm of rotation.

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Most popular questions from this chapter

A thin-walled pipe rolls along the floor. What is the ratio of its translational kinetic energy to its rotational kinetic energy about the central axis parallel to its length?

A disk with a rotational inertia of \(7.00 \mathrm{~kg} \cdot \mathrm{m}^{2}\) rotates like a merry-go-round while undergoing a time-dependent torque given by \(\tau=(5.00+2.00 t) \mathrm{N} \cdot \mathrm{m} .\) At time \(t=1.00 \mathrm{~s}\), its angular momentum is \(5.00 \mathrm{~kg} \cdot \mathrm{m}^{2} / \mathrm{s}\). What is its angular momentum at \(t=3.00 \mathrm{~s}\) ?

A yo-yo has a rotational inertia of \(950 \mathrm{~g} \cdot \mathrm{cm}^{2}\) and a mass of \(120 \mathrm{~g} .\) Its axle radius is \(3.2 \mathrm{~mm}\), and its string is \(120 \mathrm{~cm}\) long. The yo-yo rolls from rest down to the end of the string. (a) What is the magnitude of its linear acceleration? (b) How long does it take to reach the end of the string? As it reaches the end of the string, what are its (c) linear speed, (d) translational kinetic energy, (e) rotational kinetic energy, and (f) angular speed?

A uniform wheel of mass \(10.0 \mathrm{~kg}\) and radius \(0.400 \mathrm{~m}\) is mounted rigidly on a massless axle through its center (Fig. \(11-62\) ). The radius of the axle is \(0.200\) \(\mathrm{m}\), and the rotational inertia of the wheel-axle combination about its central axis is \(0.600 \mathrm{~kg} \cdot \mathrm{m}^{2}\). The wheel is initially at rest at the top of a surface that is inclined at angle \(\theta=\) \(30.0^{\circ}\) with the horizontal; the axle rests on the surface while the wheel extends into a groove in the surface without touching the surface. Once released, the axle rolls down along the surface smoothly and without slipping. When the wheel-axle combination has moved down the surface by \(2.00 \mathrm{~m}\), what are (a) its rotational kinetic energy and (b) its translational kinetic energy?

A cockroach of mass \(m\) lies on the rim of a uniform disk of mass \(4.00 \mathrm{~m}\) that can rotate freely about its center like a merry-goround. Initially the cockroach and disk rotate together with an angular velocity of \(0.260 \mathrm{rad} / \mathrm{s}\). Then the cockroach walks halfway to the center of the disk. (a) What then is the angular velocity of the cockroach- disk system? (b) What is the ratio \(K / K_{0}\) of the new kinetic energy of the system to its initial kinetic energy? (c) What accounts for the change in the kinetic energy?

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