Question

A solid body rotates about a stationary axis according to the law $\phi =at-b{t}^3$, where $a=6.0rad/s$ and $b=2.0rad/s^3$.

Find the mean values of the angular velocity and angular acceleration averaged over the time interval between $t=0$ and the complete stop. The sum of their magnitudes is x. Find the value of x.

Answer 1

$\omega =\frac{d\varphi }{dt}=a-3b{t}^2$

Time at which $\omega =0$ is $t_s=\sqrt{\frac{a}{3b}}$
${\omega }_{avg}=\frac{{\int }_0^{t_s}\omega dt}{{\int }_0^{t_s}dt}=\frac{{\int }_0^{t_s}(a-3b{t}^2)dt}{{\int }_0^{t_s}dt}=\frac{{|at-b{t}^3|}_0^{\left(\sqrt{\frac{a}{3b}}\right)}}{\sqrt{\frac{a}{3b}}}=\frac{\frac{a^2}{3b}-b\left(\frac{a^3}{27b^3}\right)}{\frac{a}{3b}}=a-\frac{a^2}{9b}=4rad/s$
$\alpha =-6bt$
${\alpha }_{avg}={|-\frac{3b{t}^2}{t}|}_0^{\sqrt{\frac{a}{3b}}}=-3b\left(\sqrt{\frac{a}{3b}}\right)=-\sqrt{3ab}=-6rad/s^2$

Sum is $4+6=10$