Question 304: 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 30
o
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 30 o 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: