Question

A light rigid rod of length L has a bob of mass M attached to one of its end just like a simple pendulum. Speed at the lowest point when it is inverted and released is

• A. $\sqrt { \mathrm { gL } }$
• B. $\sqrt { 2 g L }$
• C. $2 \sqrt { \mathrm { gL } }$
• D. $\sqrt { 5 g L }$

Answer 1

Answer: (C)

$2 \sqrt { \mathrm { gL } }$

Answer 2

P.E. with reference to lowest point = Mg(2L)

From conservation of energy,

$\frac { 1 } { 2 }$ Mv² = 2MgL

i.e. v = 2 $\sqrt { \mathrm { gL } }$