# Pendulum & Bullet, Understanding and Applying Angular Momentum

1. Dec 3, 2018

This took a lot of time and effort and I understand if you wish to skip past everything and just read my questions about it in the The too long didn't read summary (TL;DR) at the bottom.

1. The problem statement, all variables and given/known data

The 10-g bullet having a velocity of v = 750 m/s is fired into the edge of the 6-kg disk as shown. The disk is originally at rest. The rod AB has a mass of 3 kg.

Determine the magnitude of the angular velocity of the disk just after the bullet becomes embedded into its edge.

Calculate the angle θ the disk will swing when it stops.

Assumptions:
No air resistance
No friction
No other environmental forces or influences not mentioned in problem statement

Knowns:
vb = 750⋅m⋅s-1 (velocity of the bullet)
mb = 0.01⋅kg (mass of the bullet)
md = 6⋅kg (mass of the disk)
mr = 3⋅kg (mass of the rod)
Lr = 2⋅m (length of the rod)
rd = 0.4⋅m (radius of the disk)

Unknowns:
ω = ??? [rad s-1] (Angular velocity of the system just after the bullet embeds in the disk) [radians per second, 3 significant figures]

θ = ??? [°] (Change in the angle the pendulum makes from rest to its highest position before coming back after collision) [In degrees, 3 significant figures]

2. Relevant equations

Moment of Inertia for a slender rod about one end

Ir = (1/3)mr⋅Lr2

Moment of Inertia for a disk

Id = (1/2)md⋅rd2

Parallel Axis Theorem

Io = Ia + m⋅ro2

Conservation of Angular Momentum

HA,1 = HA,2

∑Ii⋅ω1 + ∑∫Mi⋅dt + ∑m⋅r⋅ivi,1 = ∑Ii⋅ω2 (this is probably where I'm getting confused)

3. The attempt at a solution

Find the mass moment of inertia of rod and disk system about point A

IA = Ir + Id + md⋅ro2

IA = (1/3)mr⋅L2 + (1/2)md⋅rd2+ md(Lr+ rd)2

IA = (1/3)(3⋅kg)(2⋅m)2 + (1/2)(6⋅kg)(0.4⋅m)2 + (6⋅kg)(2⋅m + 0.4⋅m)2

IA = 39.04⋅kg⋅m2

Apply conversation of angular momentum before impact to just after the impact.

HA,1 = HA,2

mb⋅vb(Lr+ rd) = [IA+mb(Lr+ rd)2]⋅ω

∴ ω = [mb⋅vb(Lr+ rd)]/[IA+mb⋅(Lr + rd)2]

ω = [(0.01⋅kg)(750⋅m⋅s-1)(2⋅m + 0.4⋅m)]/[(39.04⋅kg⋅m2) + (0.01⋅kg)(2⋅m + 0.4⋅m)2]

The confusion (part 1)

This part seems intuitive and correct to me (finding ω). It's also the right answer. The next part, where θ (Change in the angle the pendulum makes from rest to its highest position before coming back after collision) however I am confused as to how the procedure is intuitive or even "sensical"

What Chegg and other solution manual/websites do is "apply angular momentum after the impact"

I would probably have used conservation of energy or something of the sort but I want to know how to do it with the conservation of angular momentum because I'm not sure I fully understand how to do it and it's important for other applications plus it's the only way to get the answer they're looking for on the homework and multiple choice no-partial-credit exam.

So they start off saying something I would of figured

"Apply conservation of angular momentum after the impact"

HA,1 = HA,2

So far so good. The angular momentum of the system needs to be the same as it is right after the bullet embeds into the disk and the system starts moving with angular velocity ω .. (which is HA,1), as compared to the angular momentum it does when the pendulum has reached its max sway (HA,2).

Now the part that sends me off the deep end of what is going on and how it makes any sense what so ever

They say

-Wd(Lr + r) - Wr(1/2)L + (1/2)IAω2 = [-Wd(Lr + r) - Wr(1/2)L]cos(θ)

None of this is angular momentum? This is kinetic energy and potential energy.

So I'm wondering how they came up with this equation because even when I do conservation of energy I wind up with

Relevant Equations (part 2)

Conservation of Energy

E1 = E2

Kinetic energy (linear)

KE = (1/2)m⋅v2

Kinetic energy (angular)

KE = (1/2)I⋅ω2

Potential energy (gravitational)

PE = m⋅g⋅h

The attempt at the solution (part 2)

Esys,1 = Esys,2

KEsys,1 + PEsys,1 = KEsys,2 + PEsys,2

(1/2)Isys⋅ω12 + msys⋅g⋅h1 = (1/2)Isys⋅ω22 + msys⋅g⋅h2

ω2 will be zero since the max height will have decelerated the angular velocity completely.

h1 will be zero since this is the initial potential energy and will be referenced as zero and h2 will be the increase in vertical height when the pendulum reaches its highest point. Δh will be used to describe the change in gravitational potential energy.

Consequently the gravitational potential energy term will drop out of the left hand side of the equation or first state and the kinetic energy equation will drop out of the right hand side of the equation or second state.

(1/2)Isys⋅ω2 = msys⋅g⋅Δh2

Since we are interested in the angle θ and not Δh, Δh must be rewritten in terms of Lr, rd and θ

L being the distance from the center of mass of the part of the system to the axis o located at point a.

Applying Δh = L(1 - cos(θ)) to our energy balance equation

(1/2)Isys⋅ω2 = [md(Lr+rd) + mr(1/2)Lr]g(1-cos(θ))

θ = arccos(1 - [(1/2)Isys⋅ω2]/[md(Lr+rd) + mr(1/2)Lr]g])

θ = arccos(1 - [(1/2)(IA + mb(Lr + rd)22]/[md(Lr+rd) + mr(1/2)Lr]g])

θ = arccos(1 - [(1/2)((39.04⋅kg⋅m2) + (0.01⋅kg)(2⋅m + 0.4⋅m)2)((0.460386315 rad s-1)2]/[(6⋅kg)(2⋅m + 0.4⋅m) + (3⋅kg)(1/2)(2⋅m)](9.81⋅m⋅s-2)])

θ ≈ 12.650008408°

θ ≅ 12.7°

The confusion (part 2)

This is the correct answer and is the same value produced by the solution manuals energy balance equation yet a few major questions remain:

The too long didn't read summary (TL;DR)

1) How do you do this (find the angle θ) without energy balance and with momentum balance, or is it even possible given the information? (most important question)

2) How did the solution manual's author produce the equation for energy balance on intuition since I took a much different route as shown above. (not that important of a question)

3) What happened to the massive amount of kinetic energy in the bullet? (somewhat important)

Answered, thank you. The energy is lost due to heat from the inelastic collision.

If you consider the energy of the bullet to be

Eb = KEb = (1/2)mb⋅vb2

Eb = (1/2)(0.01⋅kg)(750⋅m⋅s-1)2

Eb = 2812.5⋅kg⋅m2s-2

Logically the energy is conserved since there is no friction and air resistance etc. etc.

So shouldn't the energy in the bullet, the energy in the pendulum after impact, and the energy during the highest point in the swing all be equal?

Eb = Esys,1 = Esys,2

(1/2)mb⋅vb2 = (1/2)Isys⋅ω2 = msys⋅g⋅Δh2

However

Esys,1 = (1/2)[(39.04⋅kg⋅m2) + (0.01⋅kg)(2⋅m + 0.4⋅m)2](0.460386315 rad s-1)2

Esys,14.143476833⋅kg⋅m2s-2

Esys,2 = [md(Lr+rd) + mr(1/2)Lr]g(1-cos(θ))

Esys,2 = [(6⋅kg)(2⋅m+0.4⋅m) + (3⋅kg)(1/2)(2⋅m)](9.81⋅m⋅s-2)(1 - cos(12.650008408°)

Esys,24.143427465 ⋅kg⋅m2s-2

EbEsys,1 ≈ Esys,2

2812.5⋅kg⋅m2s-2 >> 4.143476833⋅kg⋅m2s-24.143427465 ⋅kg⋅m2s-2

This would be a major problem if I had decided to set the energy balance to the bullet and the state where the pendulum is at its highest and then solve for θ. The answer would be drastically different and presumably wrong and I'm not sure why exactly

Eb = Esys,2

(1/2)mb⋅vb2 = [md(Lr+rd) + mr(1/2)Lr]g(1-cos(θ))

θ would be huge, also where is over 2800 J of energy going when we are ending up with only about 4 J ?

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Last edited: Dec 3, 2018
2. Dec 3, 2018

### Andrew Mason

Re: what happened to the bullet energy? The impact is not elastic. Most of the bullet energy is lost as heat. So you cannot use bullet energy before the collision to work it out. You have to determine the energy of the pendulum immediately after the bullet stops and determine potential energy of the pendulum + bullet as a function of the pendulum angle. From that you can determine how far the pendulum will swing.

AM

3. Dec 3, 2018

Oh I see, the energy from the collision goes towards deforming, embedding, and the heat associated with that.

Any insight on how to solve this problem with momentum conservation and not energy conservation (if possible?) or how the solution manual author came up with their energy balance equation?

4. Dec 4, 2018

### FactChecker

This brings up a significant point. The impact of a bullet on a target imparts less momentum than it caused on the person firing the bullet (due to speed decreasing in flight and the target does not experience a reaction to the gas momentum from the explosion). But almost all the energy of the explosion is imparted to the bullet. If the bullet stays within the target, the energy is all imparted to the target as though the gunpowder had exploded in the target. That energy causes all the damage -- not momentum.

Last edited: Dec 4, 2018
5. Dec 4, 2018

### Andrew Mason

Pendulum energy is conserved AFTER the bullet impact because we the only force acting on it is gravity, which is conservative (we are assuming that friction or air resistance is negligible). So you can use conservation of energy. You can work out the energy of the pendulum immediately after impact from its angular momentum (which you correctly determined).

AM

6. Dec 8, 2018

### haruspex

All conservation laws have provisos. For mechanical energy it is no non-conservative forces acting on the system; for linear momentum it is that the net of external forces is zero; for angular momentum about a chosen axis, that the net torque of external forces about that axis is zero.
For angular momentum, it is also necessary to pick a valid axis. This can be either a fixed point in an inertial frame or the mass centre of the system.

After the bullet's impact, as the pendulum moves from the vertical, there are two external forces: the reaction force from the hinge and gravity on the pendulum bob and bullet. To arrange that neither exerts a torque, we would need to choose an axis on the line of action of both forces. Sadly, that means the mass centre, which just leads to the angular momentum conservation equation 0=0, which gets us nowhere. And there won't be a fixed point where the net torque is zero because mg is constant while the radial force varies.
Thus, there cannot be a way to solve the problem purely by conservation of angular momentum.