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A constant external torque \(\tau\) acts for a very brief period \(\Delta t\) on a rotating system having moment of inertia \(I\), then (1) the angular momentum of the system will change by \(\tau \Delta t\) (2) the angular velocity of the system will change by \(\tau \Delta t / I\) (3) if the system was initially at rest, it will acquire rotational kinetic energy \((\tau \Delta t)^{2} / 2 I\) (4) the kinetic energy of the system will change by \((\tau \Delta t)^{2} / 2 I\)

Short Answer

Expert verified
All statements (1), (2), (3), and (4) are true.

Step by step solution

01

Understanding Torque and Angular Momentum

We start by recognizing that torque is related to the change in angular momentum. The relation between torque \( \tau \) and the change in angular momentum \( \Delta L \) is given by the equation: \( \Delta L = \tau \Delta t \). This equation tells us that the change in angular momentum is directly proportional to the applied torque and the duration \( \Delta t \) over which it is applied.Thus, the angular momentum of the system changes by \( \tau \Delta t \), confirming statement (1) as true.
02

Calculating Change in Angular Velocity

The change in angular velocity \( \Delta \omega \) can be determined using the relation between torque, moment of inertia \( I \), and angular momentum. Given \( \Delta L = I \Delta \omega \), we can express \( \Delta \omega \) as \( \frac{\Delta L}{I} \). Thus:\[ \Delta \omega = \frac{\tau \Delta t}{I} \]So, the angular velocity changes by \( \frac{\tau \Delta t}{I} \), which confirms statement (2) as true.
03

Finding Rotational Kinetic Energy Acquired

If the system starts from rest, the final angular velocity is \( \Delta \omega = \frac{\tau \Delta t}{I} \). The rotational kinetic energy \( K \) acquired can be calculated using the formula:\[ K = \frac{1}{2} I \omega^2 \]Substitute \( \omega = \frac{\tau \Delta t}{I} \) into the equation:\[ K = \frac{1}{2} I \left(\frac{\tau \Delta t}{I}\right)^2 = \frac{1}{2} \frac{(\tau \Delta t)^2}{I} \]Hence, the rotational kinetic energy becomes \( \frac{(\tau \Delta t)^2}{2I} \), supporting statement (3) as true.
04

Considering Change in Kinetic Energy

The change in kinetic energy when external torque is applied to a system from rest is given by the acquired rotational kinetic energy that we found in the previous step. Thus, the change in kinetic energy is also \( \frac{(\tau \Delta t)^2}{2I} \).Therefore, statement (4) is confirmed as true.

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Key Concepts

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

Torque
Torque is a central concept in understanding the dynamics of rotational movement. It's essentially the rotational equivalent of force. When you apply torque, you're twisting or rotating an object around an axis. Imagine turning a doorknob or using a wrench to loosen a bolt.
Torque is calculated as the product of the force applied and the distance from the axis of rotation, known as the lever arm. It's given by the formula:\[\tau = r \times F\]where \( \tau \) is torque, \( r \) is the lever arm, and \( F \) is the force. This equation conveys that the further a force is applied from the axis, the more effectively it can rotate the object.
When a constant external torque acts on a rotating system, it affects the system's angular momentum. This relationship is described by another simple yet powerful formula:\[\Delta L = \tau \Delta t\]Here, \( \Delta L \) represents the change in angular momentum, and \( \Delta t \) denotes the time over which the torque acts. This shows how a small duration force can induce a significant change in rotation if applied correctly. This direct proportionality makes torque a pivotal element in rotational dynamics.
Rotational Kinetics
Rotational kinetics deals with the motion related to objects that are rotating. Just as linear kinetics involves forces and velocity, rotational kinetics involves torque, angular velocity, and moment of inertia. These components play crucial roles in understanding and calculating how objects behave when they're in rotational motion.
In rotational mechanics, angular velocity \( \omega \) describes how fast something is spinning and is analogous to speed in linear movement. The moment of inertia \( I \) is similarly akin to mass but in terms of rotation, signifying an object's resistance to changes in its rotational state. The relationship between the torque, angular momentum, and moment of inertia is vital:\[\Delta \omega = \frac{\tau \Delta t}{I}\]This formula calculates how much the angular velocity changes, showing that a more significant torque or longer application time results in a faster spin. Conversely, a larger moment of inertia reduces the impact of the torque.
Rotational kinetic energy represents the energy possessed by a system in rotation. It's derived from both the angular velocity and the moment of inertia:\[K = \frac{1}{2} I \omega^2\]This formula calculates the energy based on how fast something spins. Any changes—for instance, an increase in angular velocity—affect the energy significantly.
Moment of Inertia
Moment of inertia is a fundamental concept when it comes to rotational motion. It essentially measures how much torque is needed to achieve a certain angular acceleration around an axis. Think of it as the rotational equivalent of mass in linear motion. A high moment of inertia means an object will be harder to start or stop rotating.
The formula for moment of inertia involves summing up all the small pieces of mass in an object multiplied by the square of their distance from the axis of rotation:\[I = \sum m_i r_i^2\]Here, each \( m_i \) is a mass element, and \( r_i \) is its distance from the axis. This equation displays that objects with mass elements far from the axis of rotation have higher moments of inertia.
This attribute affects rotational dynamics significantly. In the context of angular momentum and kinetic energy, the moment of inertia comes into play, as seen in the expressions for angular velocity and rotational kinetic energy. For example, the change in angular velocity due to a torque \( \tau \) is hampered by a high moment of inertia, thus making such systems more stable against perturbations.

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Most popular questions from this chapter

A sphere is released on a smooth inclined plane from the top. When it moves down, its angular momentum is (1) conserved about every point (2) conserved about the point of contact only (3) conserved about the centre of the sphere only (4) conserved about any point on a line parallel to the inclined plane and passing through the centre of the ball

A uniform ball of radius \(r\) rolls without slipping down from the top of a sphere of radius \(R\). The angular velocity of the ball when it breaks from the sphere is (1) \(\sqrt{\frac{5 g(R+r)}{17 r^{2}}}\) (2) \(\sqrt{\frac{10 g(R+r)}{17 r^{2}}}\) (3) \(\sqrt{\frac{5 g(R-r)}{10 r^{2}}}\) (4) \(\sqrt{\frac{10 g(R+r)}{7 r^{2}}}\)

A solid homogenous sphere is moving on a rough horizontal surface, partially rolling and partially sliding. During this kind of motion of the sphere (1) Total kinetic energy is conserved (2) Angular momentum of the sphere about the point of contact is conserved (3) Only the rotational kinetic energy about the centre of mass is conserved (4) Angular momentum about the centre of mass is conserved

A solid sphere, a hollow sphere and a disc, all having the same mass and radius, are placed at the top of an incline and released. The friction coefficients between the objects and the incline are same and not sufficient to allow pure rolling. The least time will be taken in reaching the bottom by (1) the solid sphere (2) the hollow sphere (3) the disc (4) all will take the same time

A uniform rod is resting freely over a smooth horizontal plane. A particle moving horizontally strikes at one end of the rod normally and gets stuck. Then (1) the momentum of the particle is shared between the particle and the rod and remains conserved (2) the angular momentum about the mid-point of the rod before and after the collision is equal (3) the angular momentum about the centre of mass of the combination before and after the collision is equal (4) the centre of mass of the rod particle system starts to move translationally with the original momentum of the particle

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