/*! 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 75 A particle of mass \(m\) uniform... [FREE SOLUTION] | 91Ó°ÊÓ

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A particle of mass \(m\) uniformly accelerates as it moves counterclockwise along the circumference of a circle of radius \(R\) : $$ \overrightarrow{\mathbf{r}}=\hat{\mathbf{i}} R \cos \theta+\hat{\mathbf{j}} R \sin \theta $$ with \(\theta=\omega_{0} t+\frac{1}{2} \alpha t^{2},\) where the constants \(\omega_{0}\) and \(\alpha\) are the initial angular velocity and angular acceleration, respectively. Determine the object's tangential acceleration \(\overrightarrow{\mathbf{a}}_{\tan }\) and determine the torque acting on the object using \((a) \vec{\tau}=\overrightarrow{\mathbf{r}} \times \overrightarrow{\mathbf{F}}\) (b) \(\vec{\tau}=I \vec{\alpha}\)

Short Answer

Expert verified
Tangential acceleration is \(\alpha R\) and torque is \(m \alpha R^2\).

Step by step solution

01

Identify the components

Given the position vector \(\overrightarrow{\mathbf{r}}=\hat{\mathbf{i}} R \cos \theta + \hat{\mathbf{j}} R \sin \theta\), where \(\theta=\omega_{0} t + \frac{1}{2} \alpha t^{2}\), we first need to determine the tangential acceleration of the particle as it moves along the circle's circumference.
02

Find velocity and acceleration vectors

First, compute the velocity by differentiating the position vector with respect to time: \[ \overrightarrow{\mathbf{v}} = \frac{d}{dt}(\hat{\mathbf{i}} R \cos \theta + \hat{\mathbf{j}} R \sin \theta) = -R \sin \theta \frac{d\theta}{dt} \hat{\mathbf{i}} + R \cos \theta \frac{d\theta}{dt} \hat{\mathbf{j}} \]Since \( \frac{d\theta}{dt} = \omega_0 + \alpha t \), the velocity becomes:\[ \overrightarrow{\mathbf{v}} = -R \sin \theta (\omega_0 + \alpha t) \hat{\mathbf{i}} + R \cos \theta (\omega_0 + \alpha t) \hat{\mathbf{j}} \].
03

Calculate tangential acceleration

To find tangential acceleration, differentiate the magnitude of velocity concerning time: \[ \overrightarrow{\mathbf{a}} = \frac{d}{dt} \left((\omega_0 + \alpha t) R \right) = \alpha R\].Hence, the tangential acceleration is \(\overrightarrow{\mathbf{a}}_{\tan} = \alpha R \hat{\mathbf{t}}\), where \(\hat{\mathbf{t}}\) is the unit tangent vector.
04

Using torque formulation (a) \(\vec{\tau} = \overrightarrow{\mathbf{r}} \times \overrightarrow{\mathbf{F}}\)

The force \(\overrightarrow{\mathbf{F}}\) causing the tangential acceleration is \(m\overrightarrow{\mathbf{a}}_{\tan}\). Therefore, \(\vec{\tau} = \overrightarrow{\mathbf{r}} \times (m \alpha R \hat{\mathbf{t}})\).Since \(\overrightarrow{\mathbf{r}}\) is perpendicular to \(\hat{\mathbf{t}}\), \(\vec{\tau} = m \alpha R^2\).
05

Using torque formulation (b) \(\vec{\tau} = I \vec{\alpha}\)

The torque can also be found using the relation \(\vec{\tau} = I \vec{\alpha}\).For a particle moving in a circle, the moment of inertia \(I\) is \(mR^2\), thus \(\vec{\tau} = m R^2 \alpha\).

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Key Concepts

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

Tangential Acceleration
In circular motion, tangential acceleration refers to the linear acceleration along the direction tangent to the circular path of an object. This acceleration is responsible for changing the speed of the particle as it moves along the circular path. To find the tangential acceleration, we first need to recognize its connection to angular acceleration, denoted as \( \alpha \).
For the particle in the given exercise, tangential acceleration is computed by differentiating the linear velocity with respect to time, leading to an expression that involves both \( \alpha \) and the radius \( R \). The result, as derived, is \( \overrightarrow{\mathbf{a}}_{\tan} = \alpha R \hat{\mathbf{t}} \).
This shows how the tangential acceleration is directly proportional to both the angular acceleration \( \alpha \) and the radius \( R \) of the circle.
Torque
Torque is a measure of how much a force acting on an object causes that object to rotate. In this exercise, torque is determined using two formulations which are both related to the rotational motion of the particle.
The first formulation uses the cross-product of the position vector \( \overrightarrow{\mathbf{r}} \) and the force \( \overrightarrow{\mathbf{F}} \):
  • \( \vec{\tau} = \overrightarrow{\mathbf{r}} \times \overrightarrow{\mathbf{F}} \)
  • Given that \( \overrightarrow{\mathbf{F}} = m \overrightarrow{\mathbf{a}}_{\tan} \), and \( \overrightarrow{\mathbf{a}}_{\tan} \) is perpendicular to \( \overrightarrow{\mathbf{r}} \), the torque simplifies to \( \vec{\tau} = m \alpha R^2 \).
The second formulation uses the moment of inertia \( I \) and angular acceleration \( \alpha \):
  • \( \vec{\tau} = I \vec{\alpha} \)
  • Substituting \( I = mR^2 \) for a particle moving in a circle gives \( \vec{\tau} = mR^2 \alpha \).
Both methods arrive at the same result, illustrating the consistency in torque determination in circular motion.
Angular Velocity
Angular velocity \( \omega \) describes how fast an object rotates or revolves relative to another point. It's about how quickly the angular position changes with time. For the particle in our exercise, the initial angular velocity is \( \omega_0 \).
In forms involving simple motion, we express angular velocity as \( \omega = \frac{d\theta}{dt} \), where \( \theta \) is the angular position. As the problem describes \( \theta = \omega_0 t + \frac{1}{2} \alpha t^2 \), differentiating this gives \( \omega = \omega_0 + \alpha t \).
Hence, the angular velocity increases linearly with time due to the angular acceleration \( \alpha \). It illustrates the transition between initial motion and a steadily increasing angular speed due to acceleration.
Angular Acceleration
Angular acceleration \( \alpha \) is the rate of change of angular velocity with respect to time. It determines how quickly an object speeds up or slows down its rotation. In this specific exercise, \( \alpha \) is a constant, meaning the particle speeds up at a steady rate.
We see its effect in the equation \( \theta = \omega_0 t + \frac{1}{2} \alpha t^2 \), showing how angular position changes with time.
It is a crucial factor affecting both tangential acceleration \( \overrightarrow{\mathbf{a}}_{\tan} = \alpha R\hat{\mathbf{t}} \) and the torque \( \vec{\tau} = mR^2 \alpha \). These relationships emphasize how central \( \alpha \) is in linking linear and rotational dynamics in problems dealing with circular motion.
Moment of Inertia
The moment of inertia \( I \) is a quantity expressing an object's tendency to resist angular acceleration. The larger the moment of inertia, the harder it is to change the angular velocity. For a particle moving in a circle, the moment of inertia is computed as \( I = mR^2 \).
This formula indicates how the mass \( m \) and the square of the radius \( R \) influence the object's inertia to change rotational speed. Since \( I \) directly impacts torque calculations, being one of the main factors in the formula \( \vec{\tau} = I \alpha \), understanding \( I \) helps in comprehending the distribution of mass in rotational motion.
The moment of inertia is analogous to mass in linear motion, serving a fundamental role in the equations governing the dynamics of rotating bodies.

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Most popular questions from this chapter

(II) A thin rod of mass \(M\) and length \(\ell\) is suspended vertically from a frictionless pivot at its upper end. A mass \(m\) of putty traveling horizontally with a speed \(v\) strikes the rod at its \(\mathrm{CM}\) and sticks there. How high does the bottom of the rod swing?

(II) Suppose our Sun eventually collapses into a white dwarf, losing about half its mass in the process, and winding up with a radius 1.0\(\%\) of its existing radius. Assuming the lost mass carries away no angular momentum, what would the Sun's new rotation rate be? (Take the Sun's current period to be about 30 days) What would be its final kinetic energy in terms of its initial kinetic energy of today?

Figure \(11-39\) shows a thin rod of mass \(M\) and length \(\ell\) resting on a frictionless table. The rod is struck at a distance \(x\) from its CM by a clay ball of mass \(m\) moving at speed \(v\). The ball sticks to the rod. (a) Determine a formula for the rotational motion of the system after the collision. (b) Graph the rotational motion of the system as a function of \(x,\) from \(x=0\) to \(x=\ell / 2,\) with values of \(M=450 \mathrm{~g}, \quad m=15 \mathrm{~g}\) \(\ell=1.20 \mathrm{~m},\) and \(v=12 \mathrm{~m} / \mathrm{s} .\) (c) Does the translational motion depend on \(x ?\) Explain.

A 70 -kg person stands on a tiny rotating platform with arms outstretched. (a) Estimate the moment of inertia of the person using the following approximations: the body (including head and legs) is a 60 -kg cylinder, 12 \(\mathrm{cm}\) in radius and 1.70 \(\mathrm{m}\) high; and each arm is a \(5.0-\mathrm{kg}\) thin rod, 60 \(\mathrm{cm}\) long, attached to the cylinder. (b) Using the same approximations, estimate the moment of inertia when the arms are at the person's sides. (c) If one rotation takes 1.5 \(\mathrm{s}\) when the person's arms are outstretched, what is the time for each rotation with arms at the sides? Ignore the moment of inertia of the lightweight platform. (d) Determine the change in kinetic energy when the arms are lifted from the sides to the horizontal position, (e) From your answer to part \((d),\) would you expect it to be harder or easier to lift your arms when rotating or when at rest?

(1) What is the angular momentum of a \(0.210-\) bg ball rotating on the cnd of a thin string in a circle of radius 1.35 \(\mathrm{m}\) at an angular speed of 10.4 \(\mathrm{rad} / \mathrm{s}\) ?

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