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Circular hoop Rotation problem

  • Thread starter awvvu
  • Start date
  • #1
188
1

Homework Statement


http://img410.imageshack.us/img410/6864/1975m2lm2.png [Broken]

The Attempt at a Solution


Could someone see if my solution is correct?

Part a:
[itex]I = M R^2[/itex] for a circular hoop.

[tex]L = \vec{r} \times \vec{p} = I \omega[/tex]

[tex]m_0 v_0 R \sin(\theta) = M R^2 \omega[/tex]

[tex]\omega = \frac{m_0 v_0 \sin(\theta)}{M R}[/tex]

Part b:
Using conservation of momentum to find the velocity [itex]v[/itex] of the dart+wheel system:
[tex]m_0 v_0 = (m_0 + M) v[/tex]
[tex]v = \frac{m_0 v_0}{m_0 + M}[/tex]

[tex]K_i = \frac{1}{2} m_0 v_0^2[/tex]

[tex]K_f = K_{translational} + K_{rotation} = \frac{1}{2}(M + m_0) v^2 + \frac{1}{2} (M + m_0) R^2 \omega^2[/tex]

And then just plug [itex]v[/itex] and [itex]\omega[/itex] in from above and calculate the ratio of final to initial? So, after a bunch of algebra:

[tex]\frac{K_f}{K_i} = m_0 \left(\frac{\sin^2(\theta)}{M}+\frac{\sin^2(\theta) m_0}{M^2}+\frac{1}{M+m_0}\right)[/tex]
 
Last edited by a moderator:

Answers and Replies

  • #2
454
0
for part a you should have used [tex] M + m_0 [/tex] instead of M as the mass of the complete system. the dart still has some angular momentum after it sticks to the now rotating wheel.

for part b I think the axle doesn't move, so [tex]K_{translational} = 0 [/tex]
 
  • #3
188
1
Yeah, you're absolutely right for both of them. Thanks.

The final answer for any future googlers is (oh wait, the problem text was in an image):

[tex]\frac{m_0 \sin^2(\theta)}{m_0 + M}[/tex]
 
Last edited:

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