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PhMichael
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A massless rod connects two similar particles of mass M and is rotating on the plane of a frictioless horizontal table with an angular velocity \omega. At some instant, a particles of mass 2M is put on the table such that it sticks to one of the M particles and the system rotates about its new center of mass. Find the new angular velocity.
2. The attempt at a solution
Angular momentum is conserved about the center of the "circle", A.
The angular momntum before the coliision is:
\vec{L}_{A} = I_{A} \vec{\omega} = -2Ma^{2} \omega \hat{z}
The new center of mass is: (where the origin is taken at the center and the y-axis is along the rod pointing to the upper mass and the x-axis is pointing to the right relative to y)
\vec{r}_{cm}=\frac{M(0,a)+3M(-a,0)}{4M}=(-0.75a,0.25a)
The angular momentum after the collision will have the form:
\vec{L'}_{A}=I_{A}\vec{\omega '}=(I_{cm}+4MD^{2}) \vec{\omega '}
where,
I_{cm} = 3M[(0.25a)^{2}+(0.25a)^{2}]+M[(a+0.75a)^{2}+(0.25a)^{2}]=3.5Ma^{2}
D^{2}=(0.75a)^{2}+(0.25a)^{2}=0.625a^{2}
Therefore,
\vec{L'}_{A}=(3.5Ma^{2}+4M(0.625a^{2})) \vec{\omega '}=-6Ma^{2} \omega ' \hat{z}
and equating \vec{L}_{A} with \vec{L'}_{A} yields:
\omega ' = \frac{\omega}{3}
However, the correct answer is:
\omega ' = \frac{2 \omega}{3}
Where is my mistake? =/
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Homework Statement
A massless rod connects two similar particles of mass M and is rotating on the plane of a frictioless horizontal table with an angular velocity \omega. At some instant, a particles of mass 2M is put on the table such that it sticks to one of the M particles and the system rotates about its new center of mass. Find the new angular velocity.
2. The attempt at a solution
Angular momentum is conserved about the center of the "circle", A.
The angular momntum before the coliision is:
\vec{L}_{A} = I_{A} \vec{\omega} = -2Ma^{2} \omega \hat{z}
The new center of mass is: (where the origin is taken at the center and the y-axis is along the rod pointing to the upper mass and the x-axis is pointing to the right relative to y)
\vec{r}_{cm}=\frac{M(0,a)+3M(-a,0)}{4M}=(-0.75a,0.25a)
The angular momentum after the collision will have the form:
\vec{L'}_{A}=I_{A}\vec{\omega '}=(I_{cm}+4MD^{2}) \vec{\omega '}
where,
I_{cm} = 3M[(0.25a)^{2}+(0.25a)^{2}]+M[(a+0.75a)^{2}+(0.25a)^{2}]=3.5Ma^{2}
D^{2}=(0.75a)^{2}+(0.25a)^{2}=0.625a^{2}
Therefore,
\vec{L'}_{A}=(3.5Ma^{2}+4M(0.625a^{2})) \vec{\omega '}=-6Ma^{2} \omega ' \hat{z}
and equating \vec{L}_{A} with \vec{L'}_{A} yields:
\omega ' = \frac{\omega}{3}
However, the correct answer is:
\omega ' = \frac{2 \omega}{3}
Where is my mistake? =/
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