Question 316: Centre of Mass and Rotational Motion

Question

A thin rod MN, free to rotate in the vertical plane aboutthe fixed end N, is held horizontal. When the end M is released the speed of this end, when the rod makes an angle $$\alpha $$ with the horizontal, will be proportional to : (see figure)

Accepted Answer

When the end M is released and rod makes angle $$\alpha$$ with horizontal, the displacement of centre of mass is $${L \over 2}\sin \alpha $$.
Now, we know $$mg{L \over 2}\sin \alpha  = {1 \over 2}I{\omega ^2}$$
Here, $$I = {{m{L^2}} \over 3}$$; therefore,
$$mg{L \over 2}\sin \alpha  = {1 \over 2}{{m{L^2}} \over 3}{\omega ^2}$$
$$ \Rightarrow {\omega ^2} = {{3g\sin \alpha } \over L} \Rightarrow \omega  = \sqrt {{{3g\sin \alpha } \over L}}  \Rightarrow \omega  \propto \sqrt {\sin \alpha } $$

Answer

$$\sqrt {\sin \alpha } $$

Answer

$${\sin \alpha }$$

Answer

$$\sqrt {\cos \alpha } $$

Answer

$${\cos \alpha }$$

Centre of Mass and Rotational Motion

A thin rod MN, free to rotate in the vertical plane aboutthe fixed end N, is held horizontal. When the end M is released the speed of this end, when the rod makes an angle $\alpha $ with the horizontal, will be proportional to : (see figure)

Choose the correct answer: