Combination of Linear and Angular momentum

  • Thread starter ness9660
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  • #1
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A solid cylinder of mass M = 42 kg, radius R = 0.14 m and uniform density is pivoted on a frictional axle coaxial with its symmetry axis. A particle of mass m = 4.2 kg and initial velocity v0 = 17 m/s (perpendicular to the cylinder's axis) flies too close to the cylinder's edge, collides with the cylinder and sticks to it. Before the collision, the cylinder was not rotating. What is its angular velocity after the collision? Answer in units of rad/s.

Im lost on forming the equation for this problem. Combination of angular momentum problems have been fairly easy using Li/Lf, I thought I would try something similiar like this:

Li=m*v for the momentum of the object, and nothing for the cylinder since it is at rest
Lf= (1/2)m1*R^2 + m2R^2

Where m1 is the mass of the cylinder and m2 that of the object.

Then as usual doing w=Li/Lf, however this produces a wrong answer. I think Im missing a key concept here in translating linear momentum to angular momentum because what Im doing is turning out very wrong.


Thanks
 

Answers and Replies

  • #2
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The initial angular momentum is [tex] mvR [/tex] assuming the the particle hits right at the top edge of the cylinder.

[tex] L = I\omega = mvR [/tex]
 
  • #3
35
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asrodan said:
The initial angular momentum is [tex] mvR [/tex] assuming the the particle hits right at the top edge of the cylinder.

[tex] L = I\omega = mvR [/tex]

Ah, thank you very much for the help.
 

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