a rod of length l rotate in a vertical have a anglular plane velocity omega and angular acceleration alpha as shown in figure at the given instant
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Answer:
the fall of centre of mass
h=
2
L
(1−cosθ)
mgh=mg
2
L
(1−cosθ)
law of conservation of energy
2
1
Iω
2
=
2
mgl
(1−cosθ)
2
1
3
mL
2
ω
2
=
2
mgL
(1−cosθ)ω
2
2L
6g
=(1−cosθ)
2L
6g
2sin
2
2
θ
ω=
L
6g
sin
2
2
θ
ω=
L
6g
sin
2
θ
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