Question

A uniform thin rod of length 0.700 m and mass 4.10 kg can rotate in a horizontal plane about a vertical axis through its center. The rod is at rest when a 3.0 g bullet traveling in the rotation plane is fired into one end of the rod. As viewed from above, the bullet's path makes angle θ = 60° with the rod. If the bullet lodges in the rod and the angular velocity of the rod is 15 rad/s immediately after the collision, what is the bullet's speed just before impact? m/s

Answer 1

Answer:

$v_o = 2761 m/s$

Explanation:

Since there is no external Force or Torque on the system of bullet and rod

So we can say that total angular momentum of the system will remain conserved about its center

So we will have

$mv_o\frac{L}{2}sin\theta = (I_{rod} + I_{bullet})\omega$

here we know that

$I_{rod} = \frac{mL^2}{12}$

$I_{rod} = \frac{4.10\times 0.70^2}{12}$

$I_{rod} = 0.167 kg m^2$

$I_{bullet} = mr^2$

$I_{bullet} = (0.003)(0.35^2)$

$I_{bullet} = 3.675 \times 10^{-4} kg m^2$

$\omega = 15 rad/s$

$\theta = 60$

now we have

$0.003(v_o)(0.35)sin60 = (0.167 + 3.675 \times 10^{-4})15$

$v_o = 2761 m/s$