EAMCET Engineering 2009 Solved Paper

In this process potential energy of the metre stick will be converted into rotational kinetic energy. PE of metre stick =‌

mgl

2

Because its centre of gravity lies at the middle of the rod.
Rotational kinetic energy E=‌

1

2

Iω2
I= moment of inertia of metre stick about point A=‌

ml2

3

.
By the law of conservation of energy
mg(‌

l

2

)‌=‌

1

2

Iω2
‌=‌

1

2

‌

ml2

3

(‌

vB

l

)2
By solving, we get vB=√3gl