Question 220: Centre of Mass and Rotational Motion

Question

A uniform rod of length ‘$l$’ is pivoted at one of its ends on a vertical shaft of negligible radius.
When the shaft rotates at angular speed $\omega $ the rod makes an angle $	heta $ with it (see figure). To find $	heta $
equate the rate of change of angular momentum (direction going into the paper) ${{m{l^2}} \over {12}}{\omega ^2}\sin 	heta \cos 	heta $
about the centre of mass (CM) to the torque provided by the horizontal and vertical forces F
H
and
F
V
about the CM. The value of $	heta $ is then such that :
![](https://imagex.cdn.examgoal.net/jgy2xukf3u5fzu/de535b3d-6241-4a0a-9077-0fb3255f0686/5639b9a0-f741-11ea-8293-3b6cc591631c/file-jgy2xukf3u5fzv-233w.png)

Accepted Answer

${F_v} = mg$
${F_H} = m{\omega ^2}{l \over 2}\sin 	heta $
$ 	herefore $ ${	au _{net}}\,about\,COM = {F_v}.{l \over 2}\sin 	heta  - {F_H}{l \over 2}\cos 	heta $
$ = {{m{l^2}} \over {12}}{\omega ^2}\sin 	heta \cos 	heta $
$ \Rightarrow $ $mg{l \over 2}\sin 	heta  - m{\omega ^2}{l \over 2}\sin 	heta {l \over 2}\cos 	heta  = {{m{l^2}} \over {12}}{\omega ^2}\sin 	heta \cos 	heta $
$ \Rightarrow {{gl} \over 2} - {{{\omega ^2}{l^2}} \over 4}\cos 	heta  = {{{l^2}} \over {12}}{\omega ^2}\cos 	heta $
$ \Rightarrow $ ${{gl} \over 2} ={\omega ^2}{l^2}\cos 	heta \left( {{1 \over {12}} + {1 \over 4}} ight)$
$ \Rightarrow $ ${{gl} \over 2} = {{{\omega ^2}{l^2}\cos 	heta } \over 3}$
$ \Rightarrow $ $\cos 	heta  = {{3g} \over {2{\omega ^2}l}}$
![](https://imagex.cdn.examgoal.net/1kjio7nan/fdaa1c08-bbac-4d39-addf-6dfa1c4f2cf7/bbf703f0-4e9a-11eb-af62-d347c4e94b2f/file-1kjio7nao-1400w.jpeg)

Answer

$\cos 	heta  = {{2g} \over {3l{\omega ^2}}}$

Answer

$\cos 	heta  = {{3g} \over {2l{\omega ^2}}}$

Answer

$\cos 	heta  = {g \over {2l{\omega ^2}}}$

Answer

$\cos 	heta  = {g \over {l{\omega ^2}}}$

Centre of Mass and Rotational Motion

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