## Possible problem errors?

stephen
Posts: 3
Joined: Sat Feb 25, 2006 10:02 am

### Possible problem errors?

Problem: Gman 8/10/03
I'm certain this problem has a mistake in it. The solution should be (20g/11L)^(1/2). The solution given forgot the 1/2 on the calculation for the kinetic energy. K= 1/2 Iw^2

Problem: forscher14 [8/17/03]
I don't understand how this problem was solved by the author. He/she says, " taking as our origin the point of contact between the road and the barrel... ...the angular momentum of the barrel is M * v * R" This doesn't make sense to me, the CM in the center of the barrel should have no angular momentum about it's point of contact with the road as they do not move in relation to each other.
I solved this problem using an energy method and assumed that none of the barrels energy was lost as heat from friction with the road.

1/2M * v^2 = 1/2 M vf^2 + 1/2 I*w^2 where I = M * R^2 w = vf / R

I ended up with vf = (1/2)^(1/2) v Did I mess up somehow?

Grant
Posts: 192
Joined: Tue May 11, 2004 7:55 pm

### Community Physics Problem #Gman 8/10/03

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Hi Stephen,

I have a few quick comments about the question I wrote. I'll try and get back to you in more detial later if this does not clear things up.

The Gman 8/10/03 problem is solved using torques about the pivot point to solve the problem. The energy equation you mention is not even used so that could not be a source of a possible error (although the problem could also be solved using energy and Lagrange's equations). The basic premise I used to solve the problem is to determine the equation of motion and then, just like a simple pendulum, quickly determine the angular frequency of small oscillations.

Perhaps somebody could also solve the problem for us all using energy and Lagrange's equations.

stephen
Posts: 3
Joined: Sat Feb 25, 2006 10:02 am
Oh I see, I solved for the angular frequency, not the angular frequency of small oscillations. This is what I should have done:
Lag = T - U =1/2 I * w^2 - m*g*L Cos (theta) D/Dt = full derivative
D/Dt (dLag/dw) - dLag/dtheata = 0 d/dx = partial derivative

Which gives I*a - m*g*L Sin(theta) = 0
From that point on it's just like your solution.

trani
Posts: 108
Joined: Sun Sep 07, 2008 1:04 am

### Re: Possible problem errors?

Hi guys, I have a question about community problem #Susan 11/26/03

I don't understand how can it be that in the limit r->R we should expect the answer to diverge. I mean if we compress the spring infinitely then we would shoot it far far far away, right? Not just drop it off the edge which would be r=R. The only thing I can think of is that in my head the definition of the compression is wrong...

I think of x as the distance i have to push the ball in (away from the edge) before releasing. Is that the wrong way to think of it?