/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 40 A constant torque acting on a un... [FREE SOLUTION] | 91Ó°ÊÓ

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A constant torque acting on a uniform circular wheel changes its angular momentum from \(A_{0}\) to \(4 A_{0}\) in \(4 \mathrm{sec}-\) onds. The magnitude of this torque is (a) \(3 A_{0} / 4\) (b) \(A_{0}\) (c) \(4 A_{\text {? }}\) (d) \(12 A_{0}\)

Short Answer

Expert verified
The torque is \(3 A_{0} / 4\) (option a).

Step by step solution

01

Understand the Relationship

The relationship between torque (\(\tau\)) and angular momentum (\(L\)) is given by the equation \(\tau = \frac{dL}{dt}\). This means the torque is the rate of change of angular momentum over time.
02

Calculate Change in Angular Momentum

Given in the problem, the angular momentum changes from \(A_{0}\) to \(4A_{0}\). Thus, the change in angular momentum \(\Delta L\) is \((4A_{0} - A_{0}) = 3A_{0}\).
03

Determine Time Interval

The problem states that the change in angular momentum occurs over \(4\) seconds. Thus, \(\Delta t = 4\, \text{seconds}\).
04

Apply the Torque Formula

Substitute the values into the relationship \(\tau = \frac{dL}{dt}\), where \(dL = \Delta L = 3A_{0}\) and \(dt = \Delta t = 4\): \[ \tau = \frac{3A_{0}}{4} \].
05

Identify the Correct Option

From the solution \(\tau = \frac{3A_{0}}{4}\), this matches option (a).

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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 quantity of rotation an object has, taking into account its mass distribution and speed of rotation. Think of it like the rotational equivalent of linear momentum. Angular momentum (L) is crucial because it is conserved in a closed system, meaning that unless an external torque is applied, the total angular momentum remains constant.
For an object rotating around an axis, the angular momentum is expressed as:\( L = I \omega \), where:
  • \( I \) is the moment of inertia, describing how mass is distributed in relation to the rotation axis.
  • \( \omega \) is the angular velocity, the speed of rotation.
When torque (\( \tau \)) is applied, it causes a change in angular momentum over time. The connection between torque and angular momentum is formulated as: \( \tau = \frac{dL}{dt} \), demonstrating that torque is the rate at which angular momentum changes. Thus, understanding angular momentum helps in predicting and analyzing rotational motion in various mechanical systems.
Rate of Change
In physics, the rate of change often refers to how a quantity changes with respect to time. Specifically, in the context of torque and angular momentum, the rate of change is the effective application of torque causing a variation in angular momentum.
From the equation \( \tau = \frac{dL}{dt} \), it becomes clear that torque drives how quickly or slowly angular momentum changes. For example, if a significant torque is applied over a short period, the angular momentum will rapidly increase, leading to a faster rotation.
Conversely, a smaller torque applied over a longer period results in a slower change in angular momentum, exhibiting gentler changes in rotation speed. This concept ensures we can predict rotational behavior in changing or maintaining rotational states across various scenarios in mechanical and engineering contexts.
Uniform Circular Motion
Uniform circular motion describes the motion of an object moving at a constant speed along a circular path. Although its speed remains constant, the direction of the object continually changes, meaning acceleration is present. This continuous change in direction implies the presence of a centripetal force, constantly pulling the object toward the center of its circular path.
In the context of the exercise, understanding uniform circular motion helps in analyzing how torque influences the rotational aspect of the wheel's motion. Even when the rotation is uniform, any applied torque can increase or decrease its angular velocity, affecting its angular momentum.
Uniform circular motion is fundamental in understanding complex systems, where components like wheels, merry-go-rounds, and celestial bodies in orbits maintain steady paths due to centripetal forces, highlighting the seamless harmony of force, motion, and velocity in maintaining equilibrium and directing movement.

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Most popular questions from this chapter

A uniform rod of length \(L\) is free to rotate in a vertical plane about a fixed horizontal axis through \(B\). The rod begins rotating from rest from its unstable equilibrium position. When it has tuned through an angle \(\theta\) its average angular velocity \(\omega\) is given as (a) \(\sqrt{\frac{6 g}{L}} \sin \theta\) (b) \(\sqrt{\frac{6 g}{L}} \sin \frac{\theta}{2}\) (c) \(\sqrt{\frac{6 g}{L}} \cos \frac{\theta}{2}\) (d) \(\sqrt{\frac{6 g}{L}} \cos \theta\)

The speed of a homogeneous solid sphere after rolling down an inclined plane of vertical height \(h\), from rest, without sliding is (a) \(\sqrt{g h}\) (b) \(\sqrt{(6 / 5) g h}\) (c) \(\sqrt{(4 / 3) g h}\) (d) \(\sqrt{(10 / 7) g h}\)

Three identical thin rods each of length \(l\) and mass \(M\) are joined together to form a letter \(H .\) The moment of inertia of the system about one of the sides of \(H\) is (a) \(\frac{M l^{2}}{3}\) (b) \(\frac{M l^{2}}{4}\) (c) \(\frac{2}{3} M l^{2}\) (d) \(\frac{4 M l^{2}}{3}\)

A particle of mass \(m\) moves with a constant velocity. Which of the following statements is not correct about its angular momentum about point \(O ?\) (a) It is zero when it is at \(A\) and moving along \(O A\). (b) It is same at all points along the line \(D E\). (c) It is of the same magnitude but oppositely directed at \(B\) and \(D\) ? (d) It increases as it moves along the line \(B C\).

A uniform solid sphere rolls on a horizontal surface at \(20 \mathrm{~m} / \mathrm{s}\). It then rolls up an incline having an angle of inclination at \(30^{\circ}\) with the horizontal. If the friction losses are negligible, the value of height \(h\) above the ground where the ball stops is (a) \(14.3 \mathrm{~m}\) (b) \(28.6 \mathrm{~m}\) (c) \(57.2 \mathrm{~m}\) (d) \(9.8 \mathrm{~m}\)

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