For the pivoted slender rod of length l as shown in figure, the angular velocity as the bar reaches the vertical position after being released in the horizontal position is $p \sqrt{\frac{6 g}{7 l}}$ . Find the value of p ___.

Open in App

Solution

$I = I_{C m} + m {(\frac{1}{4})}^{2}$
$= \frac{m l^{2}}{12} + \frac{m l^{2}}{16}$
$= \frac{7 m l^{2}}{48}$
$W e h a v e, K i + U i = K_{f} + U_{f}$
$0 + m g \frac{1}{4} = \frac{1}{2} I w^{2} + 0$
$w^{2} = \frac{m g \frac{1}{4}}{\frac{1}{2} \times \frac{7 m l^{2}}{48}}$
$W = \sqrt{\frac{24 g}{7 l}}$
$= 2 \sqrt{\frac{6 g}{7 l}}$

Suggest Corrections

Similar questions

Q. For the pivoted slender rod of length

l

as shown in figure., find the angular velocity as the bar reaches the vertical position after being released in the horizontal position.

A uniform slender bar AB of mass m and length L supported by a frictionless pivot at A is released from rest at its vertical position as shown in figure. Calculate the magnitude of the reaction at pivot when the bar just acquires the horizontal position shown dotted.