Collision between Disc and Rod: What Are the Post-Collision Velocities?

[Saitama]

Homework Statement

A small disc and a thin uniform rod of length l, whose mass is η times greater than the mass of disc, lie on a smooth horizontal plane. The disc is set in motion, in horizontal direction and perpendicular to the rod, with velocity v, after which it elastically collides with the end of the rod. Find the velocity of the disc and the angular velocity of the rod after the collision.

Homework Equations

The Attempt at a Solution

I have figured out that both linear and angular momentum are conserved here.
Conserving the angular momentum about the CM of rod,
$$mv\frac{l}{2}=Iω+mv'\frac{l}{2}$$
where
I is the moment of inertia of rod
ω is the angular velocity of rod after collision
v' is the velocity of disc after collision
Similarly, i can conserve the linear momentum and solve the problem but what bugs me here is that can i use the equation v=ωr? If i don't use this, i need to create an another equation using conservation of energy which gives me the correct answer but if i use v=ωr, i reach nowhere close to the answer. I am always confused with this, can someone clarify when can i use v=ωr and why can't i use it here?

Any help is appreciated!

 

[haruspex]

what bugs me here is that can i use the equation v=ωr? If i don't use this, i need to create an another equation using conservation of energy which gives me the correct answer but if i use v=ωr, i reach nowhere close to the answer. I am always confused with this, can someone clarify when can i use v=ωr and why can't i use it here?

You can use v=ωr to translate between v and ω if you know where the instantaneous centre of rotation is. In the present case you could find v and ω by other means, then use v=ωr to find that centre.

 

[Saitama]

You can use v=ωr to translate between v and ω if you know where the instantaneous centre of rotation is. In the present case you could find v and ω by other means, then use v=ωr to find that centre.

How should i go about finding that centre of rotation?
And why that instantaneous centre of rotation is not the CM of rod?

 

[haruspex]

How should i go about finding that centre of rotation?

You don't need to for this question. You solved it without using v=ωr.

And why that instantaneous centre of rotation is not the CM of rod?

Because it also has a linear movement. E.g. for a rolling wheel, the instantaneous centre of rotation is the point of contact of the wheel with the road.

 

[Saitama]

You don't need to for this question. You solved it without using v=ωr.

Yes, i have solved it but knowing more than way to solve the same question does no harm.

What if the question stated that the disc initially has some angular velocity too, how should i go about solving it then?

 

[haruspex]

Yes, i have solved it but knowing more than way to solve the same question does no harm.

It doesn't give you another way here. In this question, you could use v=ωr to find that centre, but you'd have to compute ω first, so you've already solved the problem.

What if the question stated that the disc initially has some angular velocity too, how should i go about solving it then?

That would only be relevant if there were friction between the disc and the rod. It would become a very messy problem.

 

[Saitama]

Sorry for asking stupid questions but how you came to know that there is a instantaneous centre of rotation here? Sorry if this is stupid but i really suck at rotational dynamics.

 

[haruspex]

Sorry for asking stupid questions but how you came to know that there is a instantaneous centre of rotation here? Sorry if this is stupid but i really suck at rotational dynamics.

For any rotating body there is an instantaneous axis of rotation. I.e. a line in space such that every part of the body is, at that instant, moving as though rotating about that line.
For a wheel rolling along a straight road, that axis is a transverse line through the point of contact of the wheel on the road.
If a uniform bar length 2L is struck at right angles on one end, its instantaneous centre of rotation will be L/3 from its centre (4L/3 from the end that was struck).
And so forth.

 

[ehild]

Sorry for asking stupid questions but how you came to know that there is a instantaneous centre of rotation here? Sorry if this is stupid but i really suck at rotational dynamics.

No need to use the instantaneous axis. I always prefer the other method: the in-plane motion of a rigid body (when all points of the body move parallel with the plane) consist of the motion of the CM and rotation about an axis through the CM and perpendicular to the plane.


For your problem with the point mass hitting the end of rod, there are three equations: one for the linear momenta, one for the angular momenta and one for the energy.

If v is the velocity of the point mass before collision, and it is perpendicular to the rod, and u is the velocity after collision (also perpendicular to the rod) and U_c is the velocity of the CM of the rod,

mv=mu+MU_c.

The angular momenta with respect to the initial centre of the rod:

mvL/2= muL/2+Iω=muL/2+(1/12)ML²ω

Conservation of energy:

mv²=mu²+MU_c ²+Iω².

Given v, you have enough equations to find u, U_c and ω.

ehild

 

[Saitama]

No need to use the instantaneous axis. I always prefer the other method: the in-plane motion of a rigid body (when all points of the body move parallel with the plane) consist of the motion of the CM and rotation about an axis through the CM and perpendicular to the plane.

I myself never found it of any use when i sit to solve the problems. Even the questions from the problem book by Irodov(the problem posted here is from the same book) never asked me to use the instantaneous axis of rotation. (Yet there is one problem which asks to find the locus of instantaneous axis of rotation.)

For your problem with the point mass hitting the end of rod, there are three equations: one for the linear momenta, one for the angular momenta and one for the energy.

If v is the velocity of the point mass before collision, and it is perpendicular to the rod, and u is the velocity after collision (also perpendicular to the rod) and U_c is the velocity of the CM of the rod,

mv=mu+MU_c.

The angular momenta with respect to the initial centre of the rod:

mvL/2= muL/2+Iω=muL/2+(1/12)ML²ω

Conservation of energy:

mv²=mu²+MU_c²+Iω².

Given v, you have enough equations to find u, U_c and ω.

ehild

I solved it exactly the same way.

 

[ehild]

Good... we are of the same kind

Sometimes the use of instantaneous axis makes the solution easier. Such is, for example, the relation between the velocity of the plank to that of the cylinder in this problem.
Or there is a spool with thread. You pull the thread. At what angles does the spool move forward and what angles does it move backwards?

Any in-plane motion of a rigid body can be considered as a single rotation about an instantaneous axis. It is easy to find: Connect the equivalent points and draw the perpendicular bisectors: they cross each other at the instantaneous axis.

ehild