A shell acquires the initial velocity v = 320 m/s, having made n = 2.0 turns inside the barrel whose length is equal to l = 2.0 m. 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.
When the shell comes out if acquires the velocity of 320 m/s.
⇒ u = 0
v = 320
s = 2 m
⇒ v2 = u2 + 2as
(320)2 = 2.a.2
⇒ a=(160)2
V = u + at
320 = (160)2 t
t = (180)sec
Now in this time the shell has rotated twice.
This means each and every point would have made one angular displacement of
θ = 4π
ω0 = 0
f = 180
θ = ω0t + 12a t2
⇒4π = 12a(180)2
a = 8π × (80)2
ωt = ω0 + a t
⇒ ωt = 8π × (80)2 × 180
= 640π
ω = 2010 radsec