A rod of length 50   c m is pivoted at one end. It is raised such that it 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 r a d   s - 1 ) will be g = 10   m s - 2

Question

A rod of length 50 &#160; c m is pivoted at one end. It is raised such that it 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 r a d &#160; s - 1 ) will be g = 10 &#160; m s - 2

Aicharya · Accepted Answer

From conservation of mechanical energy

Loss in gravitational $P E =$ gain in $K E$

$m g \frac{l}{2} \sin 30^{o} = \frac{1}{2} I ω^{2}$

$m g \frac{l}{2} \sin 30^{o} = \frac{1}{2} \frac{m l^{2}}{3} ω^{2}$

$ω = \sqrt{\frac{3 g}{2 l}} = \sqrt{\frac{3 \times 10}{2 \times 50 \times 10^{- 2}}} = \sqrt{30} r a d / s$

A rod of length 50   c m is pivoted at one end. It is raised such that it makes an angle of 30 o from…

Solution

A rod of length 50 &#160; c m is pivoted at one end. It is raised such that it makes an angle of 30 o from…

Solution

A rod of length 50 c m is pivoted at one end. It is raised such that it makes an angle of 30 o from…