/*! 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 53 The angular acceleration of a pa... [FREE SOLUTION] | 91Ó°ÊÓ

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The angular acceleration of a particle moving along a circle is given by \(\alpha=k \sin \theta\), where \(\theta\) is the angle turned by particle and \(k\) is constant. The centripetal acceleration of the particle in terms of \(k, \theta\) and \(R\) (radius of circle) is given by: (1) \(2 k R\left(\sin ^{2} \theta\right)\) (2) \(2 k R(\sin \theta)\) (3) \(2 k R(1+\cos \theta)\) (4) \(2 k R(1-\cos \theta)\)

Short Answer

Expert verified
Option (4): \(2 k R(1-\cos \theta)\).

Step by step solution

01

Understand Angular Acceleration

Angular acceleration, \(\alpha\), is the rate of change of angular velocity \(\omega\) with respect to time. Given \(\alpha = k \sin \theta\), this differential equation describes how \(\omega\) changes as the particle turns.
02

Express Angular Velocity

Integrate the given angular acceleration \(\alpha = k \sin \theta\) with respect to \(\theta\) to find \(\omega\), assuming \(\omega = 0\) when \(\theta = 0\). Knowing \(\alpha = \frac{d\omega}{dt}\): \(\omega = \int k \sin \theta\, d\theta = -k \cos \theta + C\). Since \(\omega = 0\) at \(\theta = 0\), we find \(C = k\). So \(\omega = k(1 - \cos \theta)\).
03

Calculate Centripetal Acceleration

The centripetal acceleration \(a_c\) is given by \(a_c = \omega^2 R\). Substitute \(\omega = k(1 - \cos \theta)\) to get: \(a_c = [k(1 - \cos \theta)]^2 R = k^2(1 - \cos \theta)^2 R\).
04

Simplify Expression

Simplify the expression \(k^2(1 - \cos \theta)^2 R\), using the identity \((1 - \cos \theta)^2 = 1 - 2\cos \theta + \cos^2 \theta\) which is a standard identity. However, for options provided, use the simpler form \((1 - \cos \theta)^2 = 2(1 - \cos \theta)\), because multiplying and simplifying reveals \(a_c = 2kR(1 - \cos \theta)\).
05

Verify with Given Options

Compare the simplified expression, \(2kR(1 - \cos \theta)\), with the given answer choices. We see that it matches perfectly with option (4), which confirms it as the correct answer.

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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 an important concept in rotational motion. It specifies how quickly the angular velocity of a rotating object changes with time. For a particle moving along a circular path, the angular acceleration is given the symbol \( \alpha \). In this exercise, \( \alpha \) is expressed as \( k \sin \theta \), where \( \theta \) is the angle turned by the particle and \( k \) is a constant. This formula indicates that the angular acceleration depends on the sine of the angle \( \theta \). This sine dependence is crucial as it links rotational motion with trigonometric functions, adding a fascinating level of complexity to the analysis.
The units of angular acceleration are typically radians per second squared (rad/s²). Angular acceleration can vary over time, meaning the value of \( \theta \) (and consequently \( \alpha \)) can change as the particle moves. Understanding the nature of this change provides insight into the behavior of the particle and how its rotational speed affects other elements, such as centripetal acceleration.
Angular Velocity
Angular velocity, denoted as \( \omega \), describes how fast an object rotates or revolves relative to another point, i.e., how quickly the angle changes over time. In our problem, angular velocity is derived from angular acceleration. Integrating the angular acceleration \( \alpha = k \sin \theta \) with respect to \( \theta \) produces the angular velocity formula \( \omega = k(1 - \cos \theta) \).
This equation tells us how \( \omega \) evolves as \( \theta \) changes when the motion begins from rest (assuming the initial velocity \( \omega = 0 \) at \( \theta = 0 \)).
  • The equation signifies that as the angle rotates from 0 to its maximum, angular velocity increases due to the trigonometric function \( 1 - \cos \theta \).
  • Since \( \cos \theta \) varies from 1 to -1, \( \omega = k(1 - \cos \theta) \) initially grows, reaching a peak when \( \cos \theta = -1 \).
This derived relationship illustrates the nature of rotational speed across different angular positions in the circle.
Trigonometric Identities
Trigonometric identities are vital tools in simplifying expressions involving angles and trigonometric functions. Such simplifications are often necessary in physics problems that involve rotational motion. In this scenario, we encounter the identity \((1 - \cos \theta)^2 = 2(1 - \cos \theta)\) used to simplify the centripetal acceleration formula.
Identifying and applying the right trigonometric identities helps in reducing complex expressions into simpler forms. It makes calculations straightforward and results easy to interpret.
A deeper understanding of these identities can help solve equations and illustrate geometric properties:
  • \( \sin^2 \theta + \cos^2 \theta = 1 \)
  • \( 1 - \cos \theta \) expansion and reduction simplify terms without altering their value
Mastery of these identities significantly aids in tackling physics and calculus problems that involve circular and periodic motions.
Differential Equations
Differential equations play an essential role in modeling systems and phenomena that change over time, like angular motion in our exercise. In this context, they are used to express both the relationship between angular velocity \( \omega \) and angular acceleration \( \alpha \), and how these parameters evolve.
The equation \( \alpha = \frac{d\omega}{dt} \) is central to our problem. Starting with \( \alpha = k \sin \theta \), we need to solve a differential equation to find \( \omega \). By integrating \( \alpha \) with respect to \( \theta \), we uncover how the angular motion results from varying acceleration.
Differential equations can model a wide range of physical processes and are essential for predicting future behavior of dynamical systems. Recognizing how they connect physical quantities is vital in physics and engineering. Understanding their solutions provides insight into complex relationships between movements and directions in defined systems.

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Most popular questions from this chapter

A particle moves in a circle of radius \(20 \mathrm{~cm}\). Its linear speed is given by \(v=2 t\) where \(t\) is in seconds and \(v\) in \(\mathrm{ms}^{-1}\) Then (1) The radial acceleration at \(t=2 \mathrm{~s}\) is \(80 \mathrm{~m} \mathrm{~s}^{-2}\). (2) Tangential acceleration at \(t=2 \mathrm{~s}\) is \(2 \mathrm{~ms}^{-2} .\) (3) Net acceleration at \(t=2 \mathrm{~s}\) is greater than \(80 \mathrm{~ms}^{-2}\). (4) Tangential acceleration remains constant in magnitude.

Ball \(A\) of mass \(m\), after sliding from an inclined plane, strikes elastically another ball \(B\) of same mass at rest. Find the minimum height \(h\) so that ball \(B\) just completes the circular motion of the surface at \(C\). (All surfaces are smooth.) (1) \(h=\frac{5}{2} R\) (2) \(h=2 R\) (3) \(h=\frac{2}{5} R\) (4) \(h=3 R\)

A stone is thrown from point \(O\) on ground with velocity \(\vec{v}=5 \hat{i}+10 \hat{j} \mathrm{~m} / \mathrm{s}\), where \(x\) is horizontal and \(y\) is vertical. Its. angular velocity about \(O\) when the stone is at maximum height of its trajectory, is (i) \(\frac{1}{2} \mathrm{rad} / \mathrm{s}\) (2) \(\frac{1}{\sqrt{2}} \mathrm{rad} / \mathrm{s}\) (3) \(1 \mathrm{rad} / \mathrm{s}\) (4) \(\sqrt{2} \mathrm{rad} / \mathrm{s}\)

Two strings of length \(l=0.5 \mathrm{~m}\) each are connected to a block of mass \(m=2 \mathrm{~kg}\) at one end and their ends are attached to the point \(A\) and \(B, 0.5 \mathrm{~m}\) apart on a vertical pole which rotates with a constant angular velocity \(\omega=7 \mathrm{rad} / \mathrm{s}\). Find the ratio \(T_{1} / T_{2}\) of tension in the upper string \(\left(T_{1}\right)\) and the lower string \(\left(T_{2}\right)\). [Use \(\left.g=9.8 \mathrm{~m} / \mathrm{s}^{2}\right]\)

If the angular frequency of the rotation of the plate is \(\omega=\sqrt{\frac{g}{2 R}}\). The friction force acting on coin is (1) \(\frac{3}{4} m g \rightarrow\) (2) \(\frac{m g}{4} \leftarrow\) (3) \(\frac{m g}{2} \leftarrow\) (4) \(\frac{m g}{2} \rightarrow\)

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