Imagine a mass $A = 1kg$ and a mass $B = 12kg$ are connected by a rigid, inextensible, massless rod of length $2m$. The masses and rod are in a horizontal line. Three other $1kg$ masses are similarly connected around $B$ in 3-dimensional space, such that the center of mass for the whole system lies at $B$, and the mass of the whole system is $16kg$.

The system starts at rest. $A$ is impacted by another body resulting in an instantaneous force perpendicular to the line between $A$ and $B$. Afterwards, $A$ has velocity $5ms^{-1}$ in the direction of the force.

I am led to believe this velocity will be partially due to linear motion of the whole system, and partially due to rotation of $A$ around $B$.

Is it possible to split this $5ms^{-1}$ velocity into a linear and rotational/tangential component?

If not, are there any heuristics I can apply to get a believable split between the two? (this problem is for a computer model which needs to look right, but doesn't necessarily have to be 100% accurate).

  • $\begingroup$ You need to show a bit of initial analysis, or the question might be closed as off-topic/no effort. I think you ask a good question, but show us some attempt at a solution (split or not), then edit the question. $\endgroup$
    – Bill N
    Feb 24, 2018 at 15:29
  • $\begingroup$ @BillN I previously had no idea how to solve the problem, and thus could not give a solution attempt. The best I could do was give a specific example of the general problem, which I did. Also, given that you agree it's a good question, there's no reason to close it when it already has a comprehensive accepted answer. $\endgroup$ Feb 25, 2018 at 17:01

1 Answer 1


You're right when you say that

[...] velocity will be partially due to linear motion of the whole system, and partially due to rotation of A around B.

indeed, the velocity of $A$ is the sum of two contributions: the velocity of the center of mass and the rotation velocity around the center of mass itself.

Let's call $V_{\mathrm{CM}}$ the first and $V_R=\Omega R$ the second, with $\Omega$ angular velocity of the body and $R$ the distance between the center of mass and $A$ (the center of mass lies on the rod between A and B). $R$ is just the coordinate of the center of mass with respect to A $$ R = R_{\mathrm{CM}}^A = \frac{(m_B+3m_C)l}{m_A+m_B+3m_C} $$ where $l$ is the length of the rod and the $m$s the masses. Reading the problem I would assume $V_A=5m/s$ to be the velocity right after the kick at the system, so in that instant $V_A$ it is true that $$ V_A = V_{\mathrm{CM}}+V_R = V_{\mathrm{CM}}+\Omega R $$ pointing in the direction of the kick, so it is orthogonal to the rod. However, this is one equation and two unknowns velocities, so there must be another relation between them. If the rod has been hit by an instantaneous force, than an unknown momentum $P$ has been transferred. This momentum acts both on the velocity of the center of the mass and the rotation of our object. So, first of all, we conserve the momentum $$ P = V_{\mathrm{CM}}M $$ where M is the total mass of the system. Second, this impulse will rotate the system, and we use the conservation of angular momentum. Lets write than the angular momentum $L$ around the center of mass: $$ L = I_{\mathrm{CM}}\Omega = PR $$ where $I_{\mathrm{CM}}$ is the momentum of inertia of our object with respect to the center of mass $$ I_{\mathrm{CM}} = m_AR^2+(m_B+3m_C)(l-R)^2 $$ Finally, substituting, we find that $$ \frac{I_\mathrm{CM}\Omega}{R} = V_{\mathrm{CM}}M $$ which is the other relation we need together with the one above (that involving $V_A$) to find both $V_{\mathrm{CM}}$ and $V_R$.

  • $\begingroup$ In all these statements, to be accurate, you must add in which frame of reference you are measuring the velocity. And if you "frame" the question that way, (pun intended), it becomes a simple transformation problem - transforming velocity from one frame of reference (the frame the object was fixed in before it was struck), and the frame of reference of the objects Center of Mass after it has been struck. $\endgroup$ Feb 12, 2018 at 20:18

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