Answer:
3.67 rad/s
Explanation:
L = Length of meter stick = 1 m
r = Distance at which the bullet will hit the stick = [tex]\frac{L}{4}=\frac{1}{4}[/tex]
m = Mass of bullet = 3 g
M = Mass of stick = 270 g
[tex]v_1[/tex] = Velocity of bullet = 250 m/s
[tex]v_2[/tex] = Velocity of bullet leaving = 140 m/s
Initial angular momentum
[tex]L_i=mv_1r[/tex]
Final angular momentum of the system
[tex]L_f=\frac{1}{12}ML^2\omega+mv_2r[/tex]
Since, angular momentum is conserved we have
[tex]mv_1r=\frac{1}{12}ML^2\omega+mv_2r\\\Rightarrow \omega=\frac{12(mv_1r-mv_2r)}{ML^2}\\\Rightarrow \omega=\frac{12(0.003\times 250\times \frac{1}{4}-0.003\times 140\times \frac{1}{4})}{0.27\times 1^2}\\\Rightarrow \omega=3.67\ rad/s[/tex]
The angular speed is the stick spinning after the collision is 3.67 rad/s
