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Found in: Page 287

Fundamentals Of Physics

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

The angular acceleration of a wheel is ${\mathbf{\alpha }}{\mathbf{=}}{\mathbf{6}}{\mathbf{.}}{\mathbf{0}}{{\mathbf{t}}}^{{\mathbf{4}}}{\mathbf{-}}{\mathbf{4}}{\mathbf{.}}{\mathbf{0}}{{\mathbf{t}}}^{{\mathbf{2}}}$, with ${\mathbf{\alpha }}$ is in radians per second-squared and ${\mathbf{t}}$ in seconds. At time ${\mathbf{t}}{\mathbf{=}}{\mathbf{}}{\mathbf{0}}$, the wheel has an angular velocity of ${\mathbf{+}}{\mathbf{2}}{\mathbf{.}}{\mathbf{0}}{\mathbf{\text{\hspace{0.17em}}}}{\mathbf{}}\frac{\mathbf{rad}}{\mathbf{s}\mathbf{}}$ and an angular position of ${\mathbf{+}}{\mathbf{1}}{\mathbf{.}}{\mathbf{0}}{\mathbf{}}{\mathbf{rad}}$. Write expressions for (a) the angular velocity (rad/s) and (b) the angular position (rad) as functions of time (s).

1. Angular velocity of the wheel as a function of time is $\mathrm{\omega }=1.2{\mathrm{t}}^{5}-1.33{\mathrm{t}}^{3}+2.0$
2. Angular position of the wheel as a function of time is $\mathrm{\theta }=0.2\text{}{\mathrm{t}}^{6}-0.33{\mathrm{t}}^{4}+2.0\mathrm{t}+1.0$
See the step by step solution

Step 1: Listing the given quantities

The angular acceleration of a wheel is,$\mathrm{\alpha }=\left(6.0{\mathrm{t}}^{4}-4.0{\mathrm{t}}^{2}\right)\text{\hspace{0.17em}}\frac{\text{rad}}{{\text{s}}^{2}}$

The angular velocity of a wheel at t = 0 is,${\mathrm{\omega }}_{0}=+2.0\text{\hspace{0.17em}}\frac{\text{rad}}{\text{s}}$

The angular position of a wheel at t = 0 is,${\mathrm{\theta }}_{0}=+1.0\text{\hspace{0.17em}}\text{rad}.$

Step 2: Determine the concept of angular acceleration and displacement

Find the angular velocity by taking time integral of angular acceleration and angular position by taking the time integral of angular velocity.

1. $\mathrm{\omega }={\int }^{\text{​}}\mathrm{\alpha }\text{dt}$
2. $\mathrm{\theta }={\int }^{\text{​}}\mathrm{\omega }\text{dt}$

Step 3: (a) Calculate angular velocity of wheel

Angular velocity is a time integral of angular acceleration. Hence,

$\begin{array}{c}\mathrm{\omega }={\int }^{\text{​}}\mathrm{\alpha }\mathrm{dt}\\ ={\int }^{\text{​}}\left(6.0{\mathrm{t}}^{4}-4.0{\mathrm{t}}^{2}\right)\mathrm{dt}\\ =6.0\left(\frac{{\mathrm{t}}^{5}}{5}\right)-\left(4.0\right)\left(\frac{{\mathrm{t}}^{3}}{3}\right)+{\mathrm{\omega }}_{0}\\ =1.2{\mathrm{t}}^{5}-1.33{\mathrm{t}}^{3}+2.0\end{array}$

Angular velocity of the wheel is $\mathrm{\omega }=1.2{\text{t}}^{5}-1.33{\text{t}}^{3}+2.0$.

Step 4: (b) Calculate angular position of wheel.

Angular position is a time integral of angular velocity. Hence:

$\begin{array}{c}\mathrm{\theta }={\int }^{\text{​}}\mathrm{\omega }\mathrm{dt}\\ ={\int }^{\text{​}}\left(1.2{\mathrm{t}}^{5}-1.33{\mathrm{t}}^{3}+2.0\right)\mathrm{dt}\\ =1.2\left(\frac{{\mathrm{t}}^{6}}{6}\right)-1.33\left(\frac{{\mathrm{t}}^{4}}{4}\right)+2.0\mathrm{t}+{\mathrm{\theta }}_{0}\\ =0.2{\mathrm{t}}^{6}-0.33{\mathrm{t}}^{4}+2.0\mathrm{t}+1.0\end{array}$

Therefore, angular position of the wheel is $=0.2{\mathrm{t}}^{6}-0.33{\mathrm{t}}^{4}+2.0\mathrm{t}+1.0$

Angular speed and angular position of an object is calculated from its angular acceleration.