A shell acquires the initial velocity `v=320m//s`, having made `n=2.0` turns inside the barrel whose length is equal to `l=2.0m`. Assuming that the shell moves inside the barrel with a uniform acceleration, find the angular velocity of its axial rotation at the moment when the shell escapes the barrel.

Explanation

The shell acquires a constant angular acceleration at the same time as it accelerates linearly. The two are related by (assuming both are constant)
`w/l=(beta)/(2pin)`
Where `w=` linear acceleration and `beta=` angular acceleration
Then, `omega=sqrt(2*beta 2pin)=sqrt(2*w/l(2pin)^2)`
But `v^2=2wl`, hence finally
`omega=(2pinv)/(l)`