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Q8P

Expert-verifiedFound in: Page 287

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).**

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

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

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,${\mathrm{\theta}}_{0}=+1.0\text{\hspace{0.17em}}\text{rad}.$

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

- $\mathrm{\omega}={\int}^{\text{}}\mathrm{\alpha}\text{dt}$
- $\mathrm{\theta}={\int}^{\text{}}\mathrm{\omega}\text{dt}$

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

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

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

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

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

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.

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