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Simple problem not working out correctly and it's driving me insane. Please help.

  • Thread starter tydychic
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  • #1
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Homework Statement


A 2.0kg (M) block hangs from the bottom of a 1kg (m), 1m (L) rod. The block and rod form a pendulum that swings on a frictionless pivot at the top end of the rod. A .01kg bullet (m') is fired into the block, where it sticks, causing the pendulum to swing out to a 30 degree angle. What was the speed of the bullet (v)? You can treat the wood block as a particle.


Homework Equations


Using conservation of energy: .5m'v[tex]^{2}[/tex]=(M+m+m')gh, with h=L-Lcos(30degrees)


The Attempt at a Solution


v=Square root[2(.01kg + 1kg + 2kg)(9.8m/sec[tex]^{2}[/tex])(1m-cos(30degrees))/.01kg)]=28.11m/s

However, the given answer is 3.9 x10[tex]^{2}[/tex]m/s...

I've been working on this problem for north of six hours now. I will greatly appreciate any insight anyone can give me.
 

Answers and Replies

  • #2
berkeman
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Homework Statement


A 2.0kg (M) block hangs from the bottom of a 1kg (m), 1m (L) rod. The block and rod form a pendulum that swings on a frictionless pivot at the top end of the rod. A .01kg bullet (m') is fired into the block, where it sticks, causing the pendulum to swing out to a 30 degree angle. What was the speed of the bullet (v)? You can treat the wood block as a particle.


Homework Equations


Using conservation of energy: .5m'v[tex]^{2}[/tex]=(M+m+m')gh, with h=L-Lcos(30degrees)


The Attempt at a Solution


v=Square root[2(.01kg + 1kg + 2kg)(9.8m/sec[tex]^{2}[/tex])(1m-cos(30degrees))/.01kg)]=28.11m/s

However, the given answer is 3.9 x10[tex]^{2}[/tex]m/s...

I've been working on this problem for north of six hours now. I will greatly appreciate any insight anyone can give me.
The Latex didn't render quite correctly, but it looks like you aren't keeping the speeding bullet's mass outside of the SQRT? Also, the center of mass of the pendulum arm is not down at L (but that wouldn't have given you the large error you are seeing).
 
  • #3
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Why would I need to calculate center of mass when i'm completely bypassing the moment of inertia?
 
  • #4
berkeman
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Why would I need to calculate center of mass when i'm completely bypassing the moment of inertia?
I certainly could be wrong, but you are calculating the final PE change from initial PE, right? The PE change of each piece is based on the movement of its center of mass, and the pendulum rod's mass is distributed.

I can try to fix your Latex, BTW. Did you want everything except the final bullet mass to be inside the SQRT sign?
 
  • #5
berkeman
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Sorry, I was wrong about moving the mass of the bullet out of the SQRT.

Is this what you meant to render?

[tex]\sqrt{\frac{2(.01kg + 1kg + 2kg)(9.8m/s^2)(1m-cos(30degrees))}{.01kg}[/tex] = 28.11m/s

So I would just make separate delta-heights for the bullet + big block mass, versus the pendulum rod mass' delta-height. But in the end, that's not going to change the answer much. I'll try checking the math next...
 
  • #6
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Yes, it is. The squared latex was messing with my square-root latex and it just got messed up.

Hmm...if you're doing the delta-heights, wouldn't the *delta-height* be the same for both the masses and the rod? I can see the height measurements themselves being different, but in that case wouldn't the rod's simply be .5m more?

Please forgive me if I'm asking incredibly stupid questions...I'm pretty sure my brain is deep-fried at this point.
 
  • #7
berkeman
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Yes, it is. The squared latex was messing with my square-root latex and it just got messed up.

Hmm...if you're doing the delta-heights, wouldn't the *delta-height* be the same for both the masses and the rod? I can see the height measurements themselves being different, but in that case wouldn't the rod's simply be .5m more?

Please forgive me if I'm asking incredibly stupid questions...I'm pretty sure my brain is deep-fried at this point.
I think the vertical motion of the center of mass of the rod will be 1/2 of the vertical motion of the center of mass of the block + bullet, since the bottom of the rod is moving with the mass+bullet, and the top of the rod is fixed.

Once that is fixed, I think you will still get an answer close to what you calculated. From there, I'd sanity check the initial KE of the bullet (calculate it), versus how high the mass and lever would have to be lifted to get the delta PE to equal what the bullet is losing.

If they don't match well, maybe we can't use conservation of energy for this problem? (Energy lost due to heat isn't accounted for maybe). Instead, you could try a momentum conservation argument, where the momentum of the rod+mass+bullet at t=0+ is equal to the bullet's momentum at t=0-. What kind of an answer does that give you? (you have to figure out how that t=0+ momentum translates into the height of the pendulum swing, but you may be able to use energy again for that, including the moment of inertia).
 
  • #8
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Okay, I double checked it solving for g instead of the height (I habitually check the constants) and it came out as 9.80002. So, either my book is scary wrong or I just set the problem up wrong to begin with...any brainwaves?
 
  • #9
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Btw, thank you!
 
  • #10
Doc Al
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Using conservation of energy: .5m'v[tex]^{2}[/tex]=(M+m+m')gh, with h=L-Lcos(30degrees)
You cannot use conservation of energy during the collision.

Treat the problem as having two parts:
(1) The collision of bullet with block/rod. What's conserved?
(2) The rising of the block/rod/bullet after the collision. (Here energy is conserved.)
 
  • #11
berkeman
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Thanks Doc. Sorry I didn't recognize that right off the bat. :blushing:
 
  • #12
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Angular velocity would be conserved?
 
  • #13
Doc Al
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Angular velocity would be conserved?
Close. Not angular velocity, but angular momentum.
 
  • #14
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Okay, I've seen several solutions to similar problems where they set it up mvL=Iw...where does the L on the left side come from?
 
  • #15
Doc Al
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Okay, I've seen several solutions to similar problems where they set it up mvL=Iw...where does the L on the left side come from?
The left hand side represents the angular momentum of the bullet before it hits the block. How is the angular momentum of a particle defined? (What's the axis about which angular momentum is conserved?)
 
  • #16
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It's defined L=Iw, but the bullet is traveling in a straight line (I'm so ignoring projectile trajectory when I say that) so it doesn't really have a rotational axis. It does once it hits the block, but then that wouldn't be pre-collision, so I don't see how it can. I can completely see how I=[tex]\beta[/tex]ML[tex]^{2}[/tex]*(v/L)=mvl (provided beta=1), but I just don't get how the bullet would be rotating about an axis.
 
  • #17
Doc Al
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It's defined L=Iw, but the bullet is traveling in a straight line (I'm so ignoring projectile trajectory when I say that) so it doesn't really have a rotational axis. It does once it hits the block, but then that wouldn't be pre-collision, so I don't see how it can. I can completely see how I=[tex]\beta[/tex]ML[tex]^{2}[/tex]*(v/L)=mvl (provided beta=1), but I just don't get how the bullet would be rotating about an axis.
The axis that we care about is the axis about which the rod is free to rotate. You need to compute the angular momentum about that axis before the collision. The angular momentum of a particle about some axis is defined as L = r X p.
 
  • #18
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well, r = L = 1m, so none of the values would change, so the erroneously way I calculated it before would have yielded the right answer. But that's obviously not right.
 
  • #19
Doc Al
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well, r = L = 1m, so none of the values would change, so the erroneously way I calculated it before would have yielded the right answer.
What calculation are you talking about? We're just talking about how to relate the bullet speed to the angular momentum. (The left hand side of mvL=Iw.) You need to calculate the angular momentum of the block/rod/bullet after the collision using conservation of energy. (The right hand side of mvL=Iw.)
 
  • #20
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Okay, here goes:

m'vL=I[tex]\omega[/tex], I= [tex]\frac{m}{3}[/tex]L[tex]^{2}[/tex]+(m'+M)L[tex]^{2}[/tex]

m'vL=([tex]\frac{m}{3}[/tex]+m'+M)L[tex]^{2}[/tex][tex]\omega[/tex]

m'vL=([tex]\frac{m}{3}[/tex]+m'+M)L[tex]^{2}[/tex]([tex]\frac{V}{L}[/tex])

m'vL= ([tex]\frac{m}{3}[/tex]+m'+M)LV

V=m'vL/((m/3)+m'+M)L

.5((m/3)+m'+M)[/tex](L squared)(V/L) = mg(.5L)(1-cos(theta)) + (m' +M)gL(1-cos(theta))

V=2[mg(.5L)(1-cos(theta))+(m' + M)gL(1-cos(theta))]/(((m/3)+m'+M)L)

v=2[mg(.5L)(1-cos(theta))+(m' + M)gL(1-cos(theta))]/{m'L}

v=659.226....not correct.
 
Last edited:
  • #21
Doc Al
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Okay, here goes:

m'vL=I[tex]\omega[/tex], I= [tex]\frac{m}{3}[/tex]L[tex]^{2}[/tex]+(m'+M)L[tex]^{2}[/tex]

m'vL=([tex]\frac{m}{3}[/tex]+m'+M)L[tex]^{2}[/tex][tex]\omega[/tex]
This I understand. Now you need to find ω.

m'vL=([tex]\frac{m}{3}[/tex]+m'+M)L[tex]^{2}[/tex][tex]\frac{V}{L}[/tex])
This and the rest I don't quite understand. What's V?

Realize that you need to find the rotational KE of the block/rod/bullet immediately after the collision.

(FYI: Your equations look funny since you're mixing Latex and non-Latex. If you want to do that, use inline tags: 'itex' instead of 'tex'.)
 
  • #22
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Sorry, Latex was giving me some trouble when the equations started getting complicated and I had to clean it up.
 
  • #23
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I replaced omega with V/L
 
  • #24
Doc Al
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  • #25
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I got it I got it I got it. This unit check is going to suck big time, but I got the numbers. Pardon me while I pass out from mental exhaustion.
 

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