Angular Momentum (Ballistic Pendulum w/ mass)

AI Thread Summary
The discussion revolves around solving a physics problem involving a ballistic pendulum, where a bullet embeds into a block, causing it to swing. Key concepts include conservation of angular momentum and energy, with the moment of inertia being crucial for calculations. The user expresses frustration over their calculations, particularly regarding the total moment of inertia, which they suspect may contain errors. They provide detailed equations but are unsure about the accuracy of their final velocity result. Clarifications on the moment of inertia formula and potential mistakes in calculations are sought to ensure a correct solution.
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I only have one more attempt on this question before I lost all of the points, so detailed help would be much appreciated. I understand everything conceptually (I think), but I don't know where I went wrong.

Homework Statement


A 2.3 kg wood block hangs from the bottom of a 1.3 kg, 1.3 meter long rod. The block and rod form a pendulum that swings on a frictionless pivot at the top end of the rod. A 12 g bullet is fired into the block, where it sticks, causing the pendulum to swing out to a 35 degree angle.

Homework Equations


Conservation of angular momentum
Conservation of energy
Moment of Inertia

The Attempt at a Solution


mb = .012 kg
mB = 2.3 kg
mR = 1.3 kg
L = 1.3 m
r = .65m
vi = ?

-Conservation of Angular momentum:

Ai = Af
Ai = (mb)(vi)(L) = IT(w) = Af

Where w is the final angular velocity
Where IT is the total moment of inertia of the system. Given by:

(1/3)((mR)(r)2 + (mb + mB)(L2)

So,

1: w =([mb * vi * L)/IT

-Conservation of Energy

KE = Change in PE
KE = (1/2)(IT)w2

Change in PE, treating initial position of the pendulum as PE = 0:

PE = mT * g * h

mT is the sum of the masses
Where h is the change in height, denoted by the change in the center of mass as the pendulum rotates:

center of mass = c = (mR * L + (mB + mb)*r)/mT

h = c - c*cos(35)

Equating KE and PE, then solving for w:

w2 = (mT * g * h)*2/IT

So w is the square root of all that mess.

Equating the above equation with 1 and solving for vi

vi = (IT* sqrt{ (mT * g * h)*2 / IT }) / ((mb)(L))

Which gives me something like 441. I'm really frustrated with this and I'm not sure what I did wrong. Thanks in advance.
 
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I don't get how you calculated total moment of inertia.
 
That might have been one of those "two o'clock in the morning typos".

That should say (1/3)(mR)(L2) + that second half. Formula of a rod rotated about an end summed with the moment of inertia of the bullet and block, treated as a point.

Beyond that, is there any other mistakes? Thanks for catching that.
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
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