MHTCET 13 May 2023 Shift 2 Solved Paper

Let the length of the rod AB be l and mass of the rod be m.
Moment of inertia of the uniform rod about an axis passing through one of its end V given as,
MI=‌

ml2

3

Rotational kinetic energy, (KE)‌rod ‌=‌

1

2

lω2
Initially the rod is held, vertical, so potential energy of rod is given by
(PE)i=mg‌

I

2

‌‌(∵‌

l

2

=COM)
Initial kinetic energy, (KE)i=0
The rod is allowed to fall, when it just touches the ground its potential energy becomes zero, i.e.,
(PE)f=0
Let, angular velocity of rod at that instant be ω, then kinetic energy at that instant is
(KE)f=‌

1

2

∕ω2
From conservation of energy,
‌(KE)i+(PE)i=(KE)f+(PE)f
‌⇒‌‌0+mg‌

l

2

=‌

1

2

lω2+0
‌⇒‌‌mgl=lω2
‌⇒‌‌mgl=‌

ml2

3

ω2
‌⇒‌‌ω=√‌

3g

l