Physical sense of angular momentum conservation [duplicate]

Question

If two balls of equal mass m are connected by a massless rod and if a force is applied to one of the ball normal to the line joining ball and the rod then what is the final subsequent motion of the system? And can we apply angular momentum conservation here?

Answer 1

The final motion will be a combination a translation and a rotation. First let us define what the rod is: the rod has a length of $2A$ with a mass $m$ on each end. We apply a force $F$ for a time $\Delta t$ on one of the masses, perpendicular to the rod. The change in momentum equals the force applied multiplied with the time the force was applied: $p=F*\Delta t$, now we have determined that $p=mv$ so then $$v_{CoM}=\frac{F*\Delta t}{2m}.^1$$ So our two rods now have a translational movement, they move in the direction the force applied.

However we now have to take a look at the rotational movement our rod will be having. The Torque that is applied is $L=F A \Delta t$, now as stated in the other answer $L=I \omega$ and thus we will need to determine the Intertial moment, which indicates how slow an object will start rotating. For this rod this is $$I=\sum m*r^2$$, r is the distance from the center of mass to the mass and m is the mass of that mass, which gives us $I=2mA^2$. Combining our results we obtain that $$\omega = \frac{F \Delta t}{2mA}$$, so this will be the angular rotation our object will be having. I hope this answers your question.

$^1$ CoM stands for center of mass, indicating it is the movement of the middle of this rod.

For some more reading on an introductory level I would recommend University Physics Or for more advanced reading I used Classical Mechanics by John R. Taylor

Answer 2

If only one ball receives force normal to the adjoining line, that force will impart a torque. Angular momentum applies, but it's not really "conserved" because the force is adding some (and we don't know what the force is from to balance the conservation equation).

The system will both translate in the direction of the force and spin.

The linear acceleration will merely be from F=ma -> a=F/m.

The angular acceleration will be from τ=Iα -> α=I/τ. The torque τ will be τ = r*F where r is the distance from the center of mass to where the torque is applied (if you look up the formula, you'll see a sin term that goes away here because sin(90)=1). I is the moment of inertia, which in the system you discribed is solved here (pdf). You can basically think of torque as an angular force and moment of inertia as an angular mass.