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A small, uniform disk of mass $m_1=2kg$ travels with a velocity $v_{1i}=3m s^{−1}$ on a frictionless surface in the xy-plane. It strikes the end of a flat, narrow, uniform rod of length, $l=4m$ and mass $m_2=1kg$. The moment of inertia of the rod about its centre of mass, $I_2=ml^2/12$ The collision is elastic and the disk leaves the collision with velocity $v_{1f}=2.35ms^{−1}$ . By considering the conservation of momentum and the conservation of angular momentum in the collision, find the translational and angular velocity of the rod after the collision. [The disk can be considered as a point particle.]

What I did was say $m(v_{1i}-v_{1f})r=I\omega$ and solve for $\omega$, which gave me $1.95s^{-1}$

To work out the final velocity I just said $mv_{1i}=mv_{1f}+mv_{2f}$ and solve for $v_{2f}$. which gave me $1.3ms^{-1}$

To check my answer, I checked if energy was conserved, I found that it wasn't.

$\frac12mv_{1i}^2=\frac12I\omega^2+\frac12mv_{2f}^2+\frac12mv_{1f}^2$

I got $LHS=9J$ and $RHS=8.9J$, surely these should be equal, as I didn't round at all?

Tom
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  • Hint: In what directions are $v_{1f}$ and $v_{2f}$? (You solved the system as if they are scalars, but they are not, they should be vectors with x and y components.) – Willie Wong Nov 30 '16 at 17:51
  • They are all in the $\hat{i}$ direction sorry – Tom Nov 30 '16 at 17:55
  • Have you done some approximation? Try to evaluate the RHS with all digits in your calculator – MattG88 Nov 30 '16 at 18:02
  • No approximation at all, $\omega=1.95$ exactly and $v_{2f}=1.3$ exactly – Tom Nov 30 '16 at 18:06

1 Answers1

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Just to sum up, we have the following system:

\begin{equation} \begin{cases} m_1(v_{1i}-v_{1f})\frac l2=I\omega\\m_1v_{1i}=m_1v_{1f}+m_2v_{2fG}\\\frac12m_1v_{1i}^2=\frac12I\omega^2+\frac12m_2v_{2fG}^2+\frac12m_1v_{1f}^2 \end{cases} \end{equation}

where $G$ is the center of mass of the rod. In this system we have two variables: $\omega,v_{2fG}$ that they can be obtained by the first two equations; the third equation can be satisfied or not. I think that this system is impossible with the data of the problem, so there must be something wrong in the formulation.

Please check it.

MattG88
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    Exactly. the system can be solved with $m_1, m_2, I$ and $v_{1i} = 3$ as prescribed for the remaining three variables. If you do so you get that $v_{1f} = 7/3 = 2.333\ldots$ which is close to, but not, the described value of 2.35 in the stated problem. – Willie Wong Nov 30 '16 at 20:04