Question

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

Hint- Equation of the rotation of the body is given then we can find the equations for the angular velocity and angular acceleration with the help of the differentiation. The differentiation of the equation of rotation will give angular velocity and the differentiation of angular velocity will give the angular acceleration.

Complete step-by-step answer:
Here we see that the equation of the rotation of the solid body about the stationary object is given,\[\varphi = at - b{t^3}\]-------equation (1), also the values of the $a$ and $b$ is given .
So, putting the values of the $a$ and $b$ in equation (1), we get
\[\varphi = 6t - 2{t^3}\]-----equation (2)
So now equation (2) is the equation of the rotation of the solid body.
Firstly, we will find the of the angular velocity and the angular acceleration,
We know that angular velocity,
$\omega = \dfrac{{d\varphi }}{{dt}}$
$ \Rightarrow \omega = \dfrac{d}{{dt}}(6t - 2{t^3})$
$ \Rightarrow \omega = 6 - 6{t^2}$------equation (3)
Now the angular acceleration,
$\alpha = \dfrac{{d\omega }}{{dt}}$
$\alpha = \dfrac{d}{{dt}}(6 - 6{t^2})$
$\alpha = - 12t$-----equation (4)
We have to calculate the value of mean angular velocity and mean angular acceleration averaged over the time interval between $t = 0$ and the complete stop.
This means that we have to calculate mean angular velocity and mean angular acceleration at the instant when the body stops (between $t = 0$ and $t = 1$).
Now,
\[{\varphi _{initial}} = 6t - 2{t^3}\]at $t = 0$
So, \[{\varphi _{initial}} = 0\]
Similarly putting $t = 1$, we get final value for the equation of rotation,
\[{\varphi _{final}} = 6 \times 1 - 2{(1)^3}\]
\[{\varphi _{final}} = 4rad/s\]
Now we will calculate the mean angular velocity with these values, as follows
${\omega _{mean}} = \dfrac{{{\varphi _{final}} - {\varphi _{initial}}}}{{{t_{final}} - {t_{initial}}}} = \dfrac{{4 - 0}}{{1 - 0}} = 4 rad/s $
Hence the value of the mean angular velocity averaged over the time interval between $t = 0$ and the complete stop is $4 rad/s $.
Now for calculation of the mean angular acceleration, we will go as follows
We know the equation of the angular velocity that is equation (3)
$\omega = 6 - 6{t^2}$
Now at $t = 0$, ${\omega _{initial}} = 6 rad/s $
And at $t = 1$, ${\omega _{final}} = 6 - 6 = 0$
Now the mean angular acceleration,
${\alpha _{mean}} = \dfrac{{{\omega _{final}} - {\omega _{initial}}}}{{{t_{final}} - {t_{initial}}}} = \dfrac{{0 - 6}}{{1 - 0}} = - 6 {rad}/ {s}^{2} $
Hence value of the mean angular acceleration averaged over the time interval between $t = 0$ and the complete stop is $ - 6 rad/s $ and the magnitude of this mean angular acceleration is $6 rad/s $
It is given that the sum of the magnitudes of the mean angular velocity and acceleration is $x$.
So, the value of the $x$,
$x = 4 + 6 = 10$.

Note- For getting the average velocity of any quantity we have to consider the changes in the initial and final quantities. initial and final quantities will tell us the average or mean value of any quantities after doing some mathematical operations. For example, here the division of difference of initial and final velocities by the difference of the initial and final time gives the mean acceleration.