Ballistic Problem where the rod has mass?

Question

A 2.0kg wood block hangs from the bottom of a 1.0kg, 1.0m long rod. The block and rod form a pendulum that swings on a fricitionless pivot at the top end of the rod. A 10g bullet is fired into the block, where it sticks, causing the pendulum to swing out to a 30 degree angle. What is the speed of the bullet? You can treat the wood block as a particle.

I have done a ballistic physics problem like this before but never one where the rod has mass. So how would I do this and incorporate the rod?

Answer 1

The problem is similar to the point mass on a pendulum, except that the moment of inertia will be the sum of the point mass, the block, and the rod.

The point mass has I= m*R^2
and the rod has I=(m*L^2)/3
since R=L
I=(m*R^2)/3

After collision, block will have mass 2.01 kg

Using conservation of momentum

.01*v is the momentum of the bullet prior to collision
and I*w is the momentum of the system

so
v=100*I*w

after collision, the system will have rotational kinetic energy of
.5*I*w^2

This will get converted into potential energy as
the block
2.01*g*h1
where cos(30)=(R-h1)/R
h1=R*(1-cos(30))
PE=2.6417
and the rod
1*g*h2
cos(30)=(R/2-h2)/(R/2)
h2=(R/2)*(1-cos(30))
PE=0.6571
so
.5*I*w^2=3.3
simplify
w=sqrt(2*3.3/I)

recall that v=100*I*w
v=100*I*sqrt(6.6/I)
v=100*sqrt(6.6*I)


compute I
2.01*R^2+(1*R^2)/3
simplify
R^2*(2.01+1/3)
R=1m
v=100*sqrt(6.6*(2.01+1/3))

v=393 m/s
j

Answer 2

m = 10 g
Mb = 2kg
Mr = 1 kg
Lr = 1 m
φ = 30
g = 9.8 m/s²

When bullet hits the rod and sticks:

Energy is NOT conserved, becuse the force of friction
which stopped the bullet converted some kinetic energy
into friction.

Momentum is NOT conserved, because of back action
force at the pivot point.

Angular momentum around the pivot point IS conserved,
because momentum pivot back action has zero lever
length, and the force of gravity is both negligible during short
time of impact and have initial zero lever too.

AFTER collision, however, angular momentum is no longer
coserved due to the force of gravity, but energy IS conserved from now on.

(A) Angular momentum before collison:
L1 = mv Lr

(B) Angular mementum after collison:
L2 = Jω = (J(rod) + J(block) + J(bullet))ω
L2 = (Mr Lr²/3 + Mb Lr² + m Lr²)ω = Lr²/3 (Mr + 3Mb + 3m)ω

(C) Conservation of alngular momentum:
L1 = L2
mv Lr = Jω = Lr²/3 (Mr + 3Mb + 3m) ω

(D) Energy after collison:
KE2 = Jω²/2 = L2²/(2J) = 3/2 (mv Lr)² / [Lr² (Mr + 3Mb + 3m)]
KE2 = Jω²/2 = L2²/(2J) = mv²/2 m/(1/3Mr + Mb + m)

(E) Potential energy at maximum swing:
PE3 = g Lr (1-cosφ) [1/2 Mr + Mb + m]
PE3 = 2g Lr sin²(φ/2) [1/2 Mr + Mb + m]

(F) Conservation of energy:
KE2 = PE3
mv²/2 m/(1/3Mr + Mb + m) = 2g Lr sin²(φ/2) [1/2 Mr + Mb + m]
v² m/(1/3Mr + Mb + m) = 4g Lr sin²(φ/2) [1/2 Mr + Mb + m]/m



Answer:
v = 2√(g Lr) sin(φ/2) √(1/3 μr + μb + 1)√(1/2 μr + μb + 1)
where
μr = Mr / m = 200
μb = Mb / m = 100

v = 2 x 3.13 m/s x 0.259 x 183 = 300 m/s