First slide

Rotational motion

Question

A particle of mass m is attached at the end B of uniform rod of same mass m and length $l$ . If the rod is whirled in vertical circle about end A, what is minimum speed of particle at end B, required to move it in vertical circle?

Moderate

A

$\sqrt{6 g l}$
B

$\sqrt{5 g l}$
C

$\sqrt{4 g l}$
D

$\sqrt{4.5 g l}$

Solution

For minimum speed supplied at bottom , at highest point $ω$ of rod should just become zero.
Gain in PE = Loss in KE
PE of Rod + PE of mass m=Rotational KE of rod+ KE of mass m
$\begin{array}{l} \Rightarrow m g (2 l) + m g l = \frac{1}{2} I ω^{2} + \frac{1}{2} m u^{2} \\ \Rightarrow m g (2 l) + m g l = \frac{1}{2} \frac{m l^{2}}{3} {(\frac{u}{l})}^{2} + \frac{1}{2} m u^{2} (∵ I = \frac{m l^{2}}{3}) \end{array}$
$\Rightarrow u = \sqrt{\frac{9 g l}{2}} = \sqrt{4.5 g l}$

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