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 $\mathbf{+}\mathbf{2}\mathbf{.}\mathbf{0}\mathbf{}\frac{\mathbf{rad}}{\mathbf{s}\mathbf{}}$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).

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

The angular velocity of a wheel at t = 0 is,${\mathrm{\omega }}_0=+2.0\frac{\text{rad}}{\text{s}}$

The angular position of a wheel at t= 0 is,${\mathrm{\theta }}_0=+1.0\text{rad}.$

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

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 ofthewheel 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.