# Mechanics problem - angular velocity

• Mar 2nd 2009, 02:10 PM
jackiemoon
Mechanics problem - angular velocity
Hi,

Can anybody help with this question please?

Two particles of mass m are joined by a thin rigid rod of negligible mass and length l. They lie on a frictionless plane and a third particle of mass m, travelling with speed V0 hits one of the two particles in the way depicted in the attached figure.

Compute the angular velocity of the rod about its mid point after the collision -
a) if the collision is elastic, so that the third mass recoils and moves straight back;
b) if the collision is completely inelastic, i.e. if the two particles colliding get stuck together.

Thanks
• Mar 3rd 2009, 01:49 AM
Momentum (and energy)
Hello jackiemoon
Quote:

Originally Posted by jackiemoon
Hi,

Can anybody help with this question please?

Two particles of mass m are joined by a thin rigid rod of negligible mass and length l. They lie on a frictionless plane and a third particle of mass m, travelling with speed V0 hits one of the two particles in the way depicted in the attached figure.

Compute the angular velocity of the rod about its mid point after the collision -
a) if the collision is elastic, so that the third mass recoils and moves straight back;
b) if the collision is completely inelastic, i.e. if the two particles colliding get stuck together.

Thanks

First let me say that it is incorrect, and misleading, to talk about the angular velocity of the rod 'about its mid point'. The angular velocity of the rod is simply the angular velocity of the rod. It's not about any particular point. Why? If at a certain instant the rod makes an angle $\displaystyle \theta$ with a fixed line in the plane, then the angular velocity of the rod is the rate of change of $\displaystyle \theta$ with respect to time.

You can calculate the angular velocity of the rod as follows:

• find the linear velocity of one point on the rod relative to another point on the rod
• then find the component of this velocity at right angles to the rod
• then divide this component by the distance between these two points.

One of these two points may be the centre of the rod and the other one end of the rod; or it may be easier to take the points at opposite ends of the rod, and ignore the centre altogether.

So, what principles can you use to solve your problem? I shall call the moving particle A, the particle on the rod that is struck B, and the other end of the rod C. Let's deal with (a) first: elastic collision. You can say:

• At the moment of impact, there is no external impulse. So the (linear) momentum of the system along and perpendicular to the line AB is unchanged.
• The impulse on A is along the line BA, so A bounces back along BA, the line along which it came. Call this rebound velocity $\displaystyle v_1$.
• The impulse on C is along the line of the rod, so C moves along the line BC.
• Split the velocity of B into two components: one along the rod ($\displaystyle v_2$) and one perpendicular to the rod ($\displaystyle v_3$). Then the velocity of C = $\displaystyle v_2$, because it's a rigid rod.
• Finally, there is no loss of kinetic energy if the collision is perfectly elastic.

These will give you three equations, from which you can find, in terms of $\displaystyle v_0$, the velocity of A, and the two components of the velocity of B (and hence the velocity of C). I make these, respectively:

$\displaystyle v_1=\frac{v_0}{7}, v_2 = \frac{2\sqrt{2}v_0}{7}, v_3 = \frac{4\sqrt{2}v_0}{7}$

Bearing in mind what I said initially, the velocity of B relative to C is therefore $\displaystyle \frac{4\sqrt{2}v_0}{7}$ at right angles to the rod, and the angular velocity of the rod is therefore $\displaystyle \frac{4\sqrt{2}v_0}{7l}$

In part (b), after impact, A and B coalesce to form a particle with mass $\displaystyle 2m$. So let the components of velocity of this new particle again be $\displaystyle v_2$ and $\displaystyle v_3$, with C's velocity, as before, being $\displaystyle v_2$. We can no longer use the KE equation, but the two components of momentum of the system along and at right angles to AB are still unchanged at impact.

So this time you get two equations, which give (according to my working):

$\displaystyle v_2 = \frac{\sqrt{2}v_0}{9}, v_3 = \frac{\sqrt{2}v_0}{6}$

and the angular velocity this time is $\displaystyle \frac{\sqrt{2}v_0}{6l}$