Question 224: 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 FH
 and
FV
 about the CM. The value of $$	heta $$ is then such that :

Accepted Answer

$${F_v} = mg$$
$${F_H} = m{\omega ^2}{l \over 2}\sin 	heta $$
$$ 	herefore $$ $${	au _{net}}\,about\,COM = {F_v}.{l \over 2}\sin 	heta  - {F_H}{l \over 2}\cos 	heta $$
$$ = {{m{l^2}} \over {12}}{\omega ^2}\sin 	heta \cos 	heta $$
$$ \Rightarrow $$ $$mg{l \over 2}\sin 	heta  - m{\omega ^2}{l \over 2}\sin 	heta {l \over 2}\cos 	heta  = {{m{l^2}} \over {12}}{\omega ^2}\sin 	heta \cos 	heta $$
$$ \Rightarrow {{gl} \over 2} - {{{\omega ^2}{l^2}} \over 4}\cos 	heta  = {{{l^2}} \over {12}}{\omega ^2}\cos 	heta $$
$$ \Rightarrow $$ $${{gl} \over 2} ={\omega ^2}{l^2}\cos 	heta \left( {{1 \over {12}} + {1 \over 4}} ight)$$
$$ \Rightarrow $$ $${{gl} \over 2} = {{{\omega ^2}{l^2}\cos 	heta } \over 3}$$
$$ \Rightarrow $$ $$\cos 	heta  = {{3g} \over {2{\omega ^2}l}}$$

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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