A uniform 1.8 m long rod with a mass of 1.20 kg is suspended vertically from a f
ID: 1378629 • Letter: A
Question
A uniform 1.8 m long rod with a mass of 1.20 kg is suspended vertically from a frictionless pin at its upper end. A 4 gram bullet with a speed of 260 m/s strikes the rod 1.3 m below the pivot point.
Determine how high the bottom of the rod swings above the datum if the bullet goes through the rod and leaves with a speed of 135 m/s.
(include units with answer)
Use COE for the swinging rod after the collision to find the change in height.
These are important values to calculate along the way. I can check them if need be
Hint #1:Use COE for the swinging rod after the collision to find the change in height.
Explanation / Answer
moment of inertia of rod about pivot, I = M*L^2/3
= 1.2*1.8^2/3
= 1.296 kg.m^2
let w is the angular speed of rod after the collsion.
Apply consrvation of angular momentum
m*u*r = m*v*r + I*w
==> w = m*r*(u-v)/I
= 0.004*1.3*(260-135)/1.296
= 0.5 rad/s
let h is the height raise by the center of mass.
after the collsion apply energy conservation
M*g*h = 0.5*I*w^2
h = 0.5*I*w^2/(M*g)
= 0.5*1.296*0.5^2/(1.2*9.8)
= 0.0138 m or 1.38 cm <<<<<<-----Answer
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