Solving 8.62: Frictionless Stick in Morin's Classical Mechanics

[Abhishek11235]

Homework Statement

This is the problem 8.62(in screenshot) from Morin's textbook of Classical mechanics. I solved it using conservation of momentum in y direction. However in solution manual,he neglects the momentum in y direction by calling stick frictionless. What is this frictionless stick and how does it help me in neglecting y-component of linear momentum?

Homework Equations

The Attempt at a Solution

 

[Anachronist]

My take on this: If it's a frictionless stick resting on a table, striking it on the end will cause the stick to start rotating. For the mass to be deflected in the y direction, the angular momentum imparted to the stick must equal the x-direction momentum of the mass. Therefore the y-momentum of the mass is irrelevant, you just need to solve for k such that the x-direction momentum of the mass equals the angular momentum of the stick.

 

[haruspex]

the angular momentum imparted to the stick must equal the x-direction momentum of the mass

An angular momentum cannot equal a linear momentum. They are dimensionally different.

 

[jbriggs444]

What is this frictionless stick and how does it help me in neglecting y-component of linear momentum?

Since the stick is frictionless, the collision can involve no force in the y direction. Only in the x direction. That means that the y component of the velocity of the mass is unchanged by the collision. And the y component of the velocity of the center of mass of the stick is also unchanged.

If you are writing down simultaneous equations (you should) that means there are fewer variables to solve for and fewer equations needed.

Ordinarily in a collision one has a huge normal force for a brief time. Then one has to consider the possibility of a huge frictional force for a similarly brief time. By making the stick frictionless, that possibility is taken off the table.

 

[Anachronist]

An angular momentum cannot equal a linear momentum. They are dimensionally different.

Apologies for my sloppiness. Yes, of course that is true, but there is a relation between the two. This post is a homework problem so I wasn't inclined to solve it -- and the question was about why the y momentum isn't considered in the solution.

 

[haruspex]

Since the stick is frictionless, the collision can involve no force in the y direction.

Quite so, but I don’t see how it affects the answer. It would only change the y direction speeds of the two objects.

 

[haruspex]

I solved it using conservation of momentum in y direction.

I do not immediately see how that would have helped to solve it. Please post your solution.

 

[Hiero]

Quite so, but I don’t see how it affects the answer. It would only change the y direction speeds of the two objects.

Yes but the y-motion is important in the elastic-energy equation. The equation for k involves v_f / v₀ (v_f being the speed of ball after collision). The frictionless-claim gives us a value for v_f / v₀

 

[haruspex]

but the y-motion is important in the elastic-energy equation.

Not really. The collision is in the x direction, and that's the bit that needs to be elastic. The total KE of the mass can be thought of as having x and y components, ½mv_x² and ½mv_y². The y component stays with the mass.
If we allow friction, there is a net loss of KE in the y components of linear motion, but the x and rotational components, before and after, are unchanged.

 

[Hiero]

@haruspex I agree now. I confused myself thinking of static friction (to conserve kinetic energy). I’m still confused though: how do we know if friction will be kinetic or static in such collision problems?

 

[haruspex]

@haruspex I agree now. I confused myself thinking of static friction (to conserve kinetic energy). I’m still confused though: how do we know if friction will be kinetic or static in such collision problems?

It's quite analogous to the non-impact case.
We consider forces in impacts to be arbitrarily large for arbitrarily short periods. That means the normal force is very large, so the available frictional force is too. If the normal impulse is J then the static frictional impulse is up to $\mu_sJ$. If that impulse is sufficient to make the two bodies move with the same velocity in the tangential direction then there is no slipping.

 

[neilparker62]

An angular momentum cannot equal a linear momentum. They are dimensionally different.

Doesn't one just multiply linear momentum (horizontal) of the ball by l/2 and equate to angular momentum of stick which then spins (presumably) about its centre?

 

[haruspex]

First, that doesn’t contradict what I wrote. An angular momentum can never equal a linear momentum.
Secondly, angular momentum needs a reference axis. If you are multiplying by l/2 then your axis is the initial position of the centre of the stick, so yes, the lost angular momentum of the mass about that point equals the angular momentum the stick gains about it. Since the linear motion of the stick is directly away from that point, such angular momentum is the same as about the stick's moving mass centre.

 

[neilparker62]

First, that doesn’t contradict what I wrote. An angular momentum can never equal a linear momentum.
Secondly, angular momentum needs a reference axis. If you are multiplying by l/2 then your axis is the initial position of the centre of the stick, so yes, the lost angular momentum of the mass about that point equals the angular momentum the stick gains about it. Since the linear motion of the stick is directly away from that point, such angular momentum is the same as about the stick's moving mass centre.

No intention to contradict - it was a genuine request for clarification which you have provided. Many thanks and I hope it helps the OP reach a solution to the problem.

 

[Abhishek11235]

I do not immediately see how that would have helped to solve it. Please post your solution.

There is change in velocity in y direction if I don't apply constraint. What I thought if a modified question was given which involves the motion in y direction,then I would be wrong.

 

[haruspex]

There is change in velocity in y direction if I don't apply constraint. What I thought if a modified question was given which involves the motion in y direction,then I would be wrong.

Yes, but you wrote that you solved the problem using conservation of momentum in the y direction. I agree it is conserved but I do not see how you used that to solve the problem. Indeed, as has been noted, it is irrelevant to the solution.