Auniform stick 1.0 m long with total mass of 270 g is pivoted at it's 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 140 m/s.

with what angular speed is the stick spinning after the collision?

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

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

grav force prop 1/r squared 2r all squared is 4rsquared 1/4rsquared quartered

your answer is, orbit around the nucleus