Question 308: Centre of Mass and Rotational Motion

Question

A rod of length 50 cm is pivoted at one end. It is raised such that if makes an angle of 30o  from the horizontal as shown and released from rest. Its angular speed when it passes through the horizontal (in rad s$$-$$1) will be (g = 10 ms$$-$$2)

Accepted Answer

When this rod move from initial position to final position then,
Gain in kinetic energy = loss in potential energy
$$ 	herefore $$  $${1 \over 2}I$$$$\omega $$
2
= mgh
$${1 \over 2}\left( {{{m{l^3}} \over 3}} ight)$$ $$\omega $$
2
= mg$$\left( {{l \over 2}\sin {{30}^ \circ }} ight)$$
$$ \Rightarrow $$  $$\omega $$
2
= $${{3g} \over {2l}}$$
$$ \Rightarrow $$  $$\omega $$ = $$\sqrt {{{3g} \over {2l}}} $$
$$ \Rightarrow $$  $$\omega $$ = $$\sqrt {{{3 	imes 10} \over {2 	imes 0.5}}} $$
$$ \Rightarrow $$  $$\omega $$ = $$\sqrt {30} $$ rad/sec

Answer

$$\sqrt {{{30} \over 7}} $$

Answer

$$\sqrt {30} $$

Answer

$${{\sqrt {20} } \over 3}$$

Answer

$${{\sqrt {30} } \over 2}$$

Centre of Mass and Rotational Motion

A rod of length 50 cm is pivoted at one end. It is raised such that if makes an angle of 30o from the horizontal as shown and released from rest. Its angular speed when it passes through the horizontal (in rad s$-1) will be (g = 10 ms-$2)

Choose the correct answer: