Find angular velocity

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  • #1
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The problem is problem 6 at

http://viewmorepics.myspace.com/index.cfm?fuseaction=viewImage&friendID=128765607&imageID=1460003525

A long thin rod of mass M and length L with two balls of mass M1 (same mass for both) attached is allowed to rotate about the horizontal axis shown. The bar is initially stationary. It is then hig with a piece of putty of mass M2 and speed v which sticks to one of the M1's.

a) Find the angular velocity of the system after the collision. The correct answer should be 6M2*v/ (6M1L + 3M2L + ML)

b) What angle will the sytem rotate through before coming to a stop? Assume that it must be between 180 and 270 degrees.



For part a,

Am I suppose to use the conservation of angular momentum.

IW = I_f*W_f
(1/12)ML^2 *w= (1/12)(M1+M2)*L^2*W_f

I am stuck though since I don't know how to account for v.

For part b,

The answer is 180 + arcsin(V^2/gL)

I don't know how to get to v^2/gL.
 

Answers and Replies

  • #2
OlderDan
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The rod with attached masses has a moment of inertia you can calculate and no initial angular momentum. You account for v by looking up the fundamental definition of angular momentum (for a moving particle; not the derived expression involving moments of inertia for a rigid assumbly of particles). It involves the mass and velocity and what else?
 
  • #3
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Angular momentum = m*r*v

I know that I for a rod is (1/12)ML^2

so L = Iw

then

L = (1/12)ML^2*v^2/L^2
L = (1/12)MV^2*L^2

r = 2L



If I have

IW = I_f*W_f

W_f = I/I_f * W

W_f = (1/12)ML^2/(I_f) * W

I am stuck. I don't know what to do from here.
 
  • #4
OlderDan
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Angular momentum = m*r*v

I know that I for a rod is (1/12)ML^2

so L = Iw

then

L = (1/12)ML^2*v^2/L^2
L = (1/12)MV^2*L^2

r = 2L



If I have

IW = I_f*W_f

W_f = I/I_f * W

W_f = (1/12)ML^2/(I_f) * W

I am stuck. I don't know what to do from here.
You need to fix the relationships between r and L and between v and ω. Be careful to distinguish the initial particle velociy from the velocity after the collision, and be careful about the lengths involved in the problem.
 
  • #5
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I made some mistakes. L = 2r, r = (1/2)L

w=v/r
w^2=v^2/r^2 or v^2/(1/2*L)^2 = 4v^2/L

Is M in this case 2M1 since we have two masses in the system before collision. Should the mass be 2M1 + M2 after collision?
 
  • #6
andrevdh
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For conservation of the angular momentum of the interacting system:

[tex]L_{before}=L_{after}[/tex]

therefore

[tex]L_{putty}=L_{system}[/tex]

[tex]\frac{l}{2}p_{putty} = \left(I_{rod} + I_{m_1m_2} + I_{m_1}\right) \omega[/tex]
 
  • #7
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Thanks. For part b, I still don't know where v^2/gl come from. G is an acceleration due to gravity, l is a length, and v^2 is m^2/s^2. After division, v^2/gl is just a number with no unit. However, I don't know where the term come from.
 
  • #8
andrevdh
Homework Helper
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Try and approach it along these lines

[tex]W_{torques} = \Delta K[/tex]

The system experiences two torques, [tex]\Gamma _1,\ \Gamma_{12}[/tex] which is from the weights of m1 and (m1 + m2). The torques will change as the system rotates so you need to integrate to find the work done by these.

The final kinetic energies are zero. So we are left with only the initial rotational kinetic energies of the three components of the system.
 
Last edited:
  • #9
andrevdh
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Note that from part a that

[tex]\omega _i = \frac{pl}{2 I_s}[/tex]

where [tex]I_s[/tex] is the moment of inertia of the system

also note that the change in kinetic energy of the system will be

[tex]\Delta K = -\frac{1}{2} I_s {\omega _i}^2[/tex]
 
Last edited:
  • #10
andrevdh
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Changed previous post.
 

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