a uniform rod of mass m and length l is at a distance xfrom a point particle mass m the acceleration of the particle just after the release from the initial point is
Answers
Answer:
Let v and w be the linear and angular velocities of the rod applying an impulse J. Then
Impulse, J = change in the Linear momentum
J=(mv
cm
−0)⟹v
cm
=
m
J
...........(i)
Angular Impuls=Change in angular momentum
Iω=J
2
l
(about C) (as shown in fig(i))
⟹
12
ml
2
ω=
2
Jl
[as I=
12
ml
2
]
⟹ω=
ml
6J
................(ii)
This is the angular velocity of the rod.
After the given time t, the rod will rotate through an angle
θ=ωt=
ml
6J
×
12J
πml
=
2
π
From equation (2)
∴v
p
′
=
6
l
ω=
6
l
×
ml
6J
=
m
J
(this velocity is due to angular rotation of the rod and it will be at 90
∘
to the velocity of v
cm
)
Now total velocity of Point P
V
P
=
V
cm
+v
p
′
=
m
J
+
m
J
V
p
=
2
m
J
Explanation: