91Ó°ÊÓ

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

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

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

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

1. $\mathrm{Ó\neg }={∫}^{\text{​}}\mathrm{Î\pm }\text{dt}$
   
2. $\mathrm{θ}={∫}^{\text{​}}\mathrm{Ó\neg }\text{dt}$
   

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

$\begin{array}{c}\mathrm{Ó\neg }={∫}^{\text{​}}\mathrm{Î\pm }\mathrm{dt}\\ ={∫}^{\text{​}}(6.0{\mathrm{t}}^4−4.0{\mathrm{t}}^2)\mathrm{dt}\\ =6.0\left(\frac{{\mathrm{t}}^5}{5}\right)−(4.0)\left(\frac{{\mathrm{t}}^3}{3}\right)+{\mathrm{Ó\neg }}_0\\ =1.2{\mathrm{t}}^5−1.33{\mathrm{t}}^3+2.0\end{array}$

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

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

$\begin{array}{c}\mathrm{θ}={∫}^{\text{​}}\mathrm{Ó\neg }\mathrm{dt}\\ ={∫}^{\text{​}}(1.2{\mathrm{t}}^5−1.33{\mathrm{t}}^3+2.0)\mathrm{dt}\\ =1.2\left(\frac{{\mathrm{t}}^6}{6}\right)−1.33\left(\frac{{\mathrm{t}}^4}{4}\right)+2.0\mathrm{t}+{\mathrm{θ}}_0\\ =0.2{\mathrm{t}}^6−0.33{\mathrm{t}}^4+2.0\mathrm{t}+1.0\end{array}$

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.