Answer:
ω = 3.66 rad/s
Explanation:
Given that
L= 1 m
M = 270 g = 0.27 kg
m = 3 g
u= 250 m/s
v= 140 m/s
We know that moment of inertia of the rod about it center
[tex]I=\dfrac{1}{12}ML^2[/tex]
[tex]I=\dfrac{1}{12}0.27\times 1^2[/tex]
I=0.0225 kg.m²
The initial angular momentum
L₁ = m u r
Final angular momentum
L₂= I ω+m v r
Here r= L/4
r= 0.25 m
There is no any external torque ,it means that angular momentum will be conserve
L₁ = L₂
m u r = I ω+m v r
m r (u - v)=I ω
Now by putting the values
m r (u - v)=I ω
0.003 x 0.25 x (250- 140)=0.0225 x ω
ω = 3.66 rad/s
