Question

A 2.4 kg block hangs from the bottom of a 2.0 kg, 1.6 m long rod. The block and the rod form a pendulum that swings out on a frictionless pivot at the top end of the rod. A 10 g bullet is fired horizontally into the block, where it sticks, causing the pendulum to swing out to an angle of 45 degrees. You can treat the wood black as a point mass. What is the moment of inertia of the pendulum about the pivot after the collision? What was the angular velocity of the bullet and block combination right after the impact? What was the initial speed of the bullet before the impact?

Answer 1

let
m1 = 2.4 kg
m2 = 2. kg
L = 1.6 m
m3 = 10 g = 0.01 kg

a) the moment of inertia of the pendulum about the pivot after the collision,

I = m1*L^2 + m2*L^2/3 + m3*L^2

= 2.4*1.6^2 + 2*1.6^2/3 + 0.01*1.6^2

= 7.88 kg.m^2 <<<<<<<<<<<<<<<-------------------Answer

b) Increase in potential energy = (m1 + m3)*g*L*(1 - cos(45)) + m2*g*(L/2)*(1 - cos(45))

= (2.4 + 0.01)*9.8*1.6*(1 - cos(45)) + 2*9.8*(1.6/2)*(1 - cos(45))

= 15.7 J

Rotational kinetic energy just after collision = Increase in potential

(1/2)*I*w^2 = delta_PE

w = sqrt(2*delta_PE/I)

= sqrt(2*15.7/7.88)

= 2.00 rad/s <<<<<<<<<<<<<<<-------------------Answer

c) let v is the speed of the bullet just before the collision.

Apply conservation of momentum

m3*v*L = I*w

v = I*w/(m3*L)

= 7.88*2/(0.01*1.6)

= 985 m/s <<<<<<<<<<<<<<<-------------------Answer