/*! 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 21 A uniform disc of mass \(M\) and... [FREE SOLUTION] | 91Ó°ÊÓ

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A uniform disc of mass \(M\) and radius \(R\) is mounted on an axle supported in frictionless bearings. A light cord is wrapped around the rim of the disc and a steady downward pull \(T\) is exerted on the cord. The angular acceleration of the disc is (1) \(\frac{T}{M R}\) (2) \(\frac{M R}{T}\) (3) \(\frac{2 T}{M R}\) (4) \(\frac{M R}{2 T}\)

Short Answer

Expert verified
The angular acceleration is \( \frac{2 T}{M R} \) (option 3).

Step by step solution

01

Understanding the Problem

We have a uniform disc of mass \(M\) and radius \(R\). A light cord is wrapped around its rim, with a force \(T\) applied downward. We need to find the angular acceleration of the disc.
02

Applying Newton's Second Law for Rotation

The torque \( \tau \) caused by the force \( T \) is given by \( \tau = T \times R \). This torque causes an angular acceleration \( \alpha \) on the disc.
03

Using the Moment of Inertia for a Disc

The moment of inertia \( I \) for a disc about an axis through its center is \( I = \frac{1}{2} M R^2 \).
04

Finding Angular Acceleration

According to the rotational form of Newton's Second Law, \( \tau = I \alpha \). Substituting the expressions for \( \tau \) and \( I \), we get: \[ T R = \frac{1}{2} M R^2 \alpha \].
05

Solving for Angular Acceleration

Rearrange the equation to solve for \( \alpha \): \[ \alpha = \frac{2 T}{M R} \].
06

Selecting the Correct Answer

Among the options given, the angular acceleration \( \alpha = \frac{2 T}{M R} \) matches option (3).

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Key Concepts

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

Angular Acceleration
Angular acceleration is a measure of how quickly an object changes its rotational speed over time. Imagine you are spinning a merry-go-round. When you push it harder, it spins faster, and that's essentially what angular acceleration describes. It's the rotational equivalent of linear acceleration.
In mathematical terms, angular acceleration, denoted by \( \alpha \), is the change in angular velocity \( \omega \) over time \( t \). It is expressed in radians per second squared \( \text{rad/s}^2 \).
For a rotational motion caused by a constant torque, the angular acceleration can be directly calculated using Newton's Second Law for Rotation:
  • \( \tau = I \alpha \)
where \( \tau \) is the torque and \( I \) is the moment of inertia. The formula effectively connects all parts of rotational dynamics together, giving us insight into how a force (via torque) and the distribution of mass (via inertia) affect the rate of change in rotational speed.
Torque
Torque is like a twist or a turning force. It is the rotational equivalent of linear force. Think of using a wrench to turn a bolt. The harder you push the wrench, and the longer the wrench is, the easier it becomes to turn the bolt. That's because both the force and the lever arm (distance to the pivot point) affect torque.
Mathematically, we express torque \( \tau \) as:
  • \( \tau = F \times r \)
where \( F \) is the force applied, and \( r \) is the perpendicular distance from the pivot point (also known as the lever arm). In the case of the disc in the exercise, the force \( T \) applied via the cord induces a torque \( \tau = T \times R \) about the center of the disc.
  • More torque means more angular acceleration.
  • The direction of torque depends on the direction of the force applied and can be clockwise or counterclockwise.
Understanding torque is crucial in situations where rotation is involved, from everyday tools to complex machines.
Moment of Inertia
The moment of inertia is like the "rotational mass" of an object. It's a measure of an object's resistance to changes in its rotational motion. Just as mass is the measure of resistance to acceleration in linear motion, the moment of inertia does the same in rotational settings.
It depends not only on the mass of the object but also on how this mass is distributed relative to the axis of rotation.
For common shapes, this is given by specific equations. For a disc about its central axis, the moment of inertia \( I \) is:
  • \( I = \frac{1}{2} M R^2 \)
where \( M \) is the mass and \( R \) is the radius of the disc.
  • This means more mass further from the axis increases the moment of inertia.
  • A higher moment of inertia means the object is harder to start or stop rotating.
Understanding the moment of inertia is key when analyzing any rotating system, as it influences how much torque is required to achieve a certain angular acceleration.

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Most popular questions from this chapter

Two rings of same radius and mass are placed such tis their centres are at a common point and their planes is perpendicular to each other. The moment of inertad the system about an axis passing through the centre perpendicular to the plane of one of the rings is (mass of ts ring \(=m\), radius \(=r)\) (1) \(\frac{1}{2} m r^{2}\) (2) \(m r^{2}\) (3) \(\frac{3}{2} m r^{2}\) (4) \(2 \mathrm{mr}^{2}\)

A sphere is moving towards the positive \(x\)-axis witn a velocity \(v_{c}\) and rotates clockwise with angular speed \(\omega\) as shown in figure, such that \(v_{c}>\omega R\). The instantaneous axis of rotation will be 1) on point \(P\) (2) on point \(P^{\prime}\) 3) inside the sphere (4) outside the sphere

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If \(I_{1}\) is the moment of inertia of a thin rod about an arig perpendicular to its length and passing through its centre of mass, and \(I_{2}\) is the moment of inertia (about central axis) of the ring formed by bending the rod, then (1) \(I_{1}: I_{2}=1: 1\) (2) \(I_{1}: I_{2}=\pi^{2}: 3\) (3) \(I_{1}: I_{2}=\pi: 4\) (4) \(I_{1}: I_{2}=3: 5\)

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