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### Fundamentals Of Physics

Book edition 10th Edition
Author(s) David Halliday
Pages 1328 pages
ISBN 9781118230718

# Figure 10-33 gives angular speed versus time for a thin rod that rotates around one end. The scale on the v axis is set by ${\mathbit{\omega }}{\mathbf{=}}{\mathbf{6}}{\mathbf{.}}{\mathbf{0}}{\mathbit{r}}{\mathbit{a}}{\mathbit{d}}{\mathbf{/}}{\mathbit{s}}$ (a) What is the magnitude of the rod’s angular acceleration? (b) At t = 4.0s , the rod has a rotational kinetic energy of 1.60J. What is its kinetic energy at t = 0 ?

1. Magnitude of the rod’s angular acceleration is, $\alpha =1.5rad/{s}^{2}$
2. The kinetic energy when t = 0 is K = 0.4J.
See the step by step solution

## Step 1: Understanding the given information

1. Angular velocity of the wheel is, ${\omega }_{2}=6.0rad/s$
2. Kinetic energy of the wheel, at t =4 s, K = 1.60 J

## Step 2: Concept and Formula used for the given question

By using the concept of Angular acceleration, Rotational kinetic energy and moment of inertia we can find angular acceleration and rotational kinetic energy at t = 0 s. The formulas used for the concept are given below.

1. Angular acceleration, ${\mathbf{\propto }}{\mathbf{=}}\frac{\mathbf{d}\mathbf{\omega }}{\mathbf{d}\mathbf{t}}$
2. Rotational Kinetic energy, ${\mathbit{K}}{\mathbf{=}}\frac{\mathbf{1}}{\mathbf{2}}{\mathbit{l}}{{\mathbit{\omega }}}^{{\mathbf{2}}}$

## Step 3: (a) Calculation for the magnitude of the rod’s angular acceleration

From the graph,

At,

${t}_{i}=0s,{\omega }_{i}=-2rad/s$

And,

${t}_{f}=4s,{\omega }_{f}=4rad/s$

Angular acceleration is the rate of change of angular velocity.

$\begin{array}{rcl}\alpha & =& \frac{d\omega }{dt}\\ & =& \frac{{\omega }_{f}-{\omega }_{i}}{{t}_{f}={t}_{i}}\end{array}$

Using all the values,

$\begin{array}{rcl}\alpha & =& \frac{4rad/s-\left(-2rad/s\right)}{4s-0}\\ & =& \frac{6}{4}rad/{s}^{2}\\ & =& 1.5rad/{s}^{2}\end{array}$

Hence the value of angular acceleration is, 1.5rad/s2.

## Step 3: (b) Calculation for the kinetic energy at t=0

Now to solve part b, we will calculate the inertia of the thin rod using rotational kinetic energy $K=1.60Jatt=4s$

Rearranging rotational kinetic energy formula for inertia I

$\begin{array}{rcl}K& =& \frac{1}{2}l{\omega }^{2}\\ l& =& \frac{2K}{{\omega }^{2}}\end{array}$

We have

$\begin{array}{rcl}att& =& 4s,K=1.60J,\omega =4rad/s\\ l& =& \frac{2×1.60J}{{\left(4rad/s\right)}^{2}}\\ & =& 0.2kg.{m}^{2}\end{array}$

Rotational inertia of the thin rod (I) is 0.2kh.m2 , we will use this to find rotational kinetic energy at,

$\begin{array}{rcl}t& =& 0s,\omega =-2rad/s\\ K& =& \frac{1}{2}l{\omega }^{2}\\ & =& \frac{1}{2}×0.2kg.{m}^{2}×{\left(-2rad/s\right)}^{2}\\ & =& 0.4J\end{array}$

Hence the kinetic energy of the thin rod at t = 0 is 0.4J.