Question 724: Centre of Mass and Rotational Motion

Question

A thin uniform rod, pivoted at O, is rotating in the horizontal plane with constant angular speed $\omega$, as shown in the figure. At time t = 0, a small insect starts from O and moves with constant speed v, with respect to the rod towards the other end. It reaches the end of the rod at t = T and stops. The angular speed of the system remains $\omega$ throughout. The magnitude of the torque (|$	au$|) about O, as a function of time is best represented by which plot ?
![](https://app-content.cdn.examgoal.net/fly/@width/image/1l24kne3x/e5d77f44-bf61-472e-8e4c-19c524999bca/f067cdd0-bf01-11ec-8fbf-67f2639a0a45/file-1l24kne3y.png?format=png)

Accepted Answer

Let M and l be the mass and length of the rod respectively and m be the mass of the insect. Let the insect be at a distance x from O at any instant of time t.
$	herefore$ x = vt ....... (i)
Angular momentum of the system about O,
$L = \left[ {{{M{l^2}} \over {12}} + m{x^2}} ight]\omega $
$ = \left[ {{{M{l^2}} \over {12}} + m{{(vt)}^2}} ight]\omega $ (Using (i))
$ = \left[ {{{M{l^2}} \over {12}} + m{v^2}{t^2}} ight]\omega $
As, $\left| {\overrightarrow 	au  } ight| = {{dL} \over {dt}} = {d \over {dt}}\left[ {{{M{l^2}} \over {12}} + m{v^2}{t^2}} ight]\omega $
As $\omega$ and v remain constant
$	herefore$ $\left| {\overrightarrow 	au  } ight| = 2m\omega {v^2}t$
$\left| {\overrightarrow 	au  } ight| \propto t$
Hence, the graph and t is a straight line passing through the (0, 0). Option (b) represents correct plot.
![](https://app-content.cdn.examgoal.net/fly/@width/image/1l253l7sl/7092c06f-d3b1-4324-80f2-e8decda1c6cb/015d1560-bf4c-11ec-b997-87cc5d16a405/file-1l253l7sm.png?format=png)

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Centre of Mass and Rotational Motion

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