Question

A solid body rotates with deceleration about a stationary axis with an angular deceleration$\beta \propto \sqrt{\omega }$, where $\omega $is its angular velocity. If the mean angular velocity of the body averaged over the whole time of rotation is $\left\langle \omega \right\rangle =\dfrac{{{\omega }_{0}}}{x}$, (at the initial moment of time, its angular velocity was equal to ${{\omega }_{0}}$) find the value of $x$.

Answer 1

Hint: This question can be solved by finding out the relation by integrating the relation of angular acceleration deceleration given with respect to time, to find out the total change in angular velocity in terms of the initial angular velocity and also the total time taken. This relation can then be used to find out the total angular rotation by integrating it with respect to time and then dividing it by the total time taken. Then we can compare this final relation that we have got with that given in the question to find out the value of $x$.
Formula used:

$\alpha =\dfrac{d\omega }{dt}$ ,
$\therefore \alpha dt=d\omega $

where $\alpha $ is the instantaneous angular acceleration of the body, $\omega $ is the angular velocity and $t$ is the time taken.

$\omega =\dfrac{d\theta }{dt}$ ,
$\therefore \omega dt=d\theta $
where $\omega $ is the instantaneous angular velocity of the body, $\theta $ is the angular rotation and $t$ is the time taken.

$\text{Mean angular velocity = }\dfrac{\text{Total angular rotation covered by a body}}{\text{time taken}}$
$\therefore \left\langle \omega \right\rangle =\dfrac{\int\limits_{0}^{\theta }{d\theta }}{\int\limits_{0}^{t}{dt}}$

Complete step by step answer:
We are given a body that is rotating about a stationary axis and has an angular deceleration $\beta \propto \sqrt{\omega }$, where $\omega $ is its angular velocity. We will first find out the change in angular velocity by integrating this relation with respect to time since,
$\alpha =\dfrac{d\omega }{dt}$ --(1)
$\therefore \alpha dt=d\omega $
where $\alpha $ is the instantaneous angular acceleration of the body, $\omega $ is the angular velocity and $t$ is the time taken.
Therefore, $\beta =K\sqrt{\omega }$--(2)
 where K is a constant of proportionality.
Using (1),
$\beta =-\dfrac{d\omega }{dt}$ --(3)

Since in the question, $\beta $ is angular deceleration, it will have a negative sign.
Now, using (2) in (3), we get,

$-\dfrac{d\omega }{dt}=K\sqrt{\omega }$
$\therefore -\dfrac{d\omega }{\sqrt{\omega }}=Kdt$
Now, integrating with proper limits and with respective variable, we get,
$-\int\limits_{{{\omega }_{0}}}^{\omega }{\dfrac{d\omega }{\sqrt{\omega }}}=K\int\limits_{0}^{T}{dt}$
where $T$ is the total time of rotation.
$\Rightarrow -2\left[ \sqrt{\omega } \right]_{{{\omega }_{0}}}^{\omega }=K\left[ t \right]_{0}^{T}$
$\Rightarrow -2\left[ \sqrt{\omega }-\sqrt{{{\omega }_{0}}} \right]=K\left[ T-0 \right]$
$\Rightarrow \sqrt{{{\omega }_{0}}}-\sqrt{\omega }=\dfrac{KT}{2}$
$\Rightarrow \sqrt{\omega }=\sqrt{{{\omega }_{0}}}-\dfrac{KT}{2}$
Now, squaring both sides, we get,
${{\left( \sqrt{\omega } \right)}^{2}}={{\left( \sqrt{{{\omega }_{0}}}-\dfrac{KT}{2} \right)}^{2}}$
$\therefore \omega ={{\omega }_{0}}-\sqrt{{{\omega }_{0}}}KT+\dfrac{{{K}^{2}}{{T}^{2}}}{4}$ ---(4)


Now, considering $\omega =0$, at the end of rotation, we get,

$2\sqrt{{{\omega }_{0}}}=KT$
$\therefore T=\dfrac{2\sqrt{{{\omega }_{0}}}}{K}$
Therefore total time of rotation $T=\dfrac{2\sqrt{{{\omega }_{0}}}}{K}$ --(5)
Now, $\text{Mean angular velocity = }\dfrac{\text{Total angular rotation covered by a body}}{\text{total time taken}}$
$\omega =\dfrac{d\theta }{dt}$ ,
$\therefore \omega dt=d\theta $--(6)

where $\omega $is the instantaneous angular velocity of the body, $\theta $ is the angular rotation and $t$ is the time taken.

We will now find out the total angular rotation $\left( \theta \right)$ using this information.
Using (6), the total angular rotation will be,

$\int\limits_{0}^{\theta }{d\theta }=\int\limits_{0}^{T}{\omega dt}$-(7)
Putting (4) in (7), we get,
\[\int\limits_{0}^{\theta }{d\theta }=\int\limits_{0}^{\dfrac{2\sqrt{{{\omega }_{0}}}}{K}}{\left( {{\omega }_{0}}-\sqrt{{{\omega }_{0}}}KT+\dfrac{{{K}^{2}}{{T}^{2}}}{4} \right)dT}\]
$\therefore \left[ \theta \right]_{0}^{\theta }=\left[ {{\omega }_{0}}T-\sqrt{{{\omega }_{0}}}K\dfrac{{{T}^{2}}}{2}+\dfrac{{{K}^{2}}}{4}\dfrac{{{T}^{3}}}{3} \right]_{0}^{\dfrac{2\sqrt{{{\omega }_{0}}}}{K}}$
$\therefore \theta =\left[ {{\omega }_{0}}\dfrac{2\sqrt{{{\omega }_{0}}}}{K}-\dfrac{\sqrt{{{\omega }_{0}}}K}{2}\dfrac{4{{\omega }_{0}}}{{{K}^{2}}}+\dfrac{{{K}^{2}}}{12}\dfrac{8{{\omega }_{0}}\sqrt{{{\omega }_{0}}}}{{{K}^{3}}}-0-0-0 \right]$
$\therefore \theta =\left[ \dfrac{24{{K}^{2}}{{\omega }_{0}}\sqrt{{{\omega }_{0}}}-24{{K}^{2}}{{\omega }_{0}}\sqrt{{{\omega }_{0}}}+8{{K}^{2}}{{\omega }_{0}}\sqrt{{{\omega }_{0}}}}{12{{K}^{3}}} \right]=\dfrac{8{{K}^{2}}{{\omega }_{0}}\sqrt{{{\omega }_{0}}}}{12{{K}^{3}}}=\dfrac{2{{\omega }_{0}}\sqrt{{{\omega }_{0}}}}{3K}$
Therefore, total angular rotation $\theta =\dfrac{2{{\omega }_{0}}\sqrt{{{\omega }_{0}}}}{3K}$ --(8)

Now, from (5), total time of rotation $T=\dfrac{2\sqrt{{{\omega }_{0}}}}{K}$.

Now, since, $\text{Mean angular velocity = }\dfrac{\text{Total angular rotation covered by a body}}{\text{total time taken}}$-(9)

Putting (8) and (5) in (9), we get,

$\left\langle \omega \right\rangle =\dfrac{\dfrac{2{{\omega }_{0}}\sqrt{{{\omega }_{0}}}}{3K}}{\dfrac{2\sqrt{{{\omega }_{0}}}}{K}}=\dfrac{{{\omega }_{0}}}{3}$
Now, comparing with the relation given in the question,
$\dfrac{{{\omega }_{0}}}{x}=\dfrac{{{\omega }_{0}}}{3}$

$\therefore x=3$

Therefore, the value of $x$ is 3.

Note: Students must take note that the angular deceleration given in the question is not constant. Many students cannot catch this point and proceed using the equations of rotational motion which are applicable only for uniform motion with constant angular acceleration (or deceleration). This leads to them solving the question completely but arriving at the wrong answer.
The correct way to solve this question is the process given above where one has to find out the instantaneous angular acceleration (which is changing with time and in this case angular velocity) and integrate it to find a relation for the instantaneous angular velocity and again integrate it with respect to time to get the total angular rotation.
If one proceeds by considering the angular acceleration to be constant, the question will be solved very easily however, the answer will be completely wrong.