Question

A uniform stick 1.0 m long with a total mass of 300.0 g is pivoted at its center. A 3.0 g bullet is shot through the stick midway between the pivot and one end. The bullet approaches at 250 m/s and leaves at 160 m/s. With what angular speed is the stick spinning after the collision, if the bullet strikes the stick perpendicularly? (The moment of inertia of a uniform stick rotating about its center is I = (1/12)mL².)

Answer 1

Step 1: Find moment of inertia of rod rotating about it's center

I = (1/12)*m*L^2

m = mass of rod = 300.0 gm = 0.300 kg

L = length of wire = 1.0 m

So, I = (1/12)*0.300*1.0^2 = 0.025 kg.m^2

Step 2: since there is no external force/torque applied on the system of bullet + rod, So angular momentum before and after collision will remain conserved, So

Li = Lf

Li = Initial angular momentum only due to translational motion of bullet = mb*Vi*(L/4)

Here L/4 = distance of bullet from pivot point (about which rod will rotate)

Lf = final angular momentum of system = L1 + L2

L1 = angular momentum due to rotational motion of rod = I*w

L2 = angular momentum of bullet after collision = mb*Vf*(L/4)

So,

mb*Vi*(L/4) = I*w + mb*Vf*L/4

w = final angular velocity = mb*L*(Vi - Vf)/(4*I)

w = 3.0*10^-3*1.0*(250 - 160)/(4*0.025)

w = 2.7 rad/sec

Let me know if you've any query.