BSc Nursing GNM 2020 Solved Paper

If the bottom end of the rod does not slip, function does not work and the bottom does not slip. Thus energy is conserved.
Centre of mass =‌

L

2

above the ground.
For vertical position
u‌‌=‌

1

2

MgL
K‌‌=0
For horizontal position
u‌‌=0,K=‌

1

2

Iω2
∴‌‌‌

1

2

MgL‌‌=‌

1

2

(‌

mL2

3

)ω2
⇒ω2‌‌=‌

3g

L

⇒ω‌‌=√‌

3g

L

⇒v‌‌=ωL=√3gL