A rod of length l is hinged from one end. Bought to horizontal position and released. Angular velocity of rod when it is in vertical position
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Answer:
The loss in gravitational potential energy of the rod's center of mass is equal to the gain in rotational kinetic energy of the rod.
Hence mg
2
L
=
2
1
Iω
2
=
2
1
(
3
mL
2
)ω
2
⟹ω=
L
3g
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