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I have a homework question in which a sticky hockey puck traveling at constant velocity parallel to the side of the rink strikes a stationary puck and sticks to it. The angels between centres at collision is 30 degrees. I must find the angular velocity of the two pucks stuck together.

I first found the center of mass velocity by the principal of conservation of momentum to be $$\frac{v}{2\cos(\alpha)}$$ where alpha is the direction of travel of the two pucks together. I am bogged down after this, partly because I am not sure if it is correct and partly because I am not sure of the most efficient method of calculating the angular velocy. By the way both pucks have mass $p$ and radius $h$. As this is homework, I would like some help in understanding and figuring this out myself rather that a straight solution.

Thanks

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  • $\begingroup$ By the way is my expression for the centre of mass velocity correct? $\endgroup$
    – user
    Feb 2, 2015 at 16:29
  • $\begingroup$ The two pucks stuck together will travel in the same direction as the original puck; so $\alpha=0$ (for other values of $\alpha$ you would need a transverse component of external force, which doesn't exist). So your answer is correct but overcomplicated. Leave the cosine term out. $\endgroup$
    – Floris
    Feb 2, 2015 at 16:33

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The angular momentum of the system is the same before and after the collision. Since one object is stationary before the collision, the angular momentum is just the momentum ($mv$ of the moving puck multiplied by the perpendicular distance between them (which is $2r\sin(30)$).

The moment of inertia of the two pucks stuck together is a little bit tricky, but necessary to solve this. You need to use the parallel axis theorem - a disk rotating about its center has moment of inertia $\frac12mr^2$, and rotating about a point on the rim you need to add another $mr^2$. The total moment of inertia of the two is therefore

$$I=2(\frac12 mr^2 + mr^2) = 3 mr^2$$

Divide angular momentum by moment of inertia, and there's your solution.

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  • $\begingroup$ Thanks for your detailed reply, I understand that angular momentum will be conserved, I have $mvr$ as the final and the initial angular momentum. But when I let angular momentum = $I.\omega$ then equate the two expressions for angular momentum and I am still left with $v$ and $\omega$ and $r$ in the equation. $\endgroup$
    – user
    Feb 2, 2015 at 16:28
  • $\begingroup$ Well - you are trying to solve for $\omega$ which is obviously going to be a function of (scale linearly with) $v$ so you expect those two to be there. And the answer will depend on the radius since moment of inertia scales with radius squared but angular momentum before the collision scales with radius. The size of a puck is well known, so you can find that. I assume that the velocity is given in the question. If it is not, then express the angular velocity in terms of the initial velocity. $\endgroup$
    – Floris
    Feb 2, 2015 at 16:31
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A key idea is that the path of the center of mass is unaffected whether the two pucks collide or not. So take the snapshot of the situation at any moment of time before collision. Find the angular momenta of the two pucks with respect to the com at that instant of time. Since angular momentum is conserved for an isolated system (i.e $\tau_{net}=0$), the angular momentum of the system will be same when the pucks stick. Therefore, the new angular velocity will be that angular momentum divided by the new moment of inertia of the system about the center of mass (which can be calculated using parallel axis theorum).

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