Mechanics + Momentum Homework: Solving Quadratic Equation

[zorro]

Homework Statement

Please refer the attachment.

Homework Equations

The Attempt at a Solution

Conserving angular momentum about the point of rotation,
mv_olcosϑ = ml²w/3 + mVl/2

Since collision is perfectly elastic, kinetic energy is conserved.
mv_o²/2 = (ml²w/3 + mV²)/2

Solving both I got a quadratic equation in w. Please tell if there is something wrong.

 

[aim1732]

Please tell me where your point of rotation is and what is V.

 

[aim1732]

Alright I get your problem.
You have wrongly assumed that the body performs pure rotation after impact. Since collision is elastic the velocity of the tip of the rod along the normal to the floor is reversed but the rod also slips along the floor because of the absence of friction. Hence it is a body executing general motion and the equations for L and energy are wrong.
One more thing-since the floor is smooth the force by the floor on the rod is perpendicular along it hence the centre of mass has its velocity along the y-axis only.

 

[zorro]

I didnot assume that it performs pure rotational motion.
mvolcosϑ = ml2w/3 + mVl/2
On the RHS I considered V as the velocity of centre of mass after collision ( i.e. the motion after collision is translatory+rotatory).
What is wrong in Equation for energy?
Can you write the correct equations?

 

[aim1732]

Well for one thing you say you conserve L about the point of rotation which by implication is the instantaneous centre-something whose position can not be determined with data given here.
So I assume it is about point of contact with the floor that you conserve L.
For a point outside the body L= I_cmw + mv_cml
I can see ml₂/3 which is moment of inertia about an end-presumably in contact with the ground.
Same with energy.
Sorry if I am getting you all wrong. These analyses are done about certain points and if not specified my and your views can differ sharply.

 

[hikaru1221]

Conserving angular momentum about the point of rotation,
mv_olcosϑ = ml²w/3 + mVl/2

Since collision is perfectly elastic, kinetic energy is conserved.
mv_o²/2 = (ml²w/3 + mV²)/2

It seems to be that you're messing up with the numbers. I'll comment on the correctness of each term first (that means, I'm not saying anything about whether your approach is correct).

_ For the 1st equation:
From the way I understand, you choose to consider the point of contact (lowest end of the rod). That's why you thought the angular momentum about that point was conserved, right? If so, then all the 3 terms are incorrect.

The LHS should be: $$\frac{1}{2}mv_olcos\theta$$ . Notice that vector $$\vec{v}_0$$ is placed in the MIDDLE of the rod.

The RHS should be: $$\frac{1}{12}ml^2\omega - \frac{1}{2}mVlcos\theta$$.
This comes from the KOENIG theorem for angular momentum: $$\vec{L}_{rod-about-A}=\vec{L}_{COM-about-A}+\vec{L}_{rod-about-COM}$$ (where A is the point of contact, COM stands for center of mass). Notice that $$\vec{V}$$ points vertically upwards and is placed in the MIDDLE of the rod. Besides, the angular momentum conservation equation is a VECTOR equation, and vectors can have different directions. When we project the vectors onto an axis, they can have either positive value or negative value! The minus sign before the 2nd term is due to that reason (did you notice that $$\vec{v}_0$$ and $$\vec{V}$$ had opposite directions? ).

_ For the 2nd equation:
The 2nd term in this equation is wrong. The moment of inertia of the rod about its COM is mL²/12. This also comes from the KOENIG theorem for kinetic energy:
$$K_{rod} = K_{COM} + K_{rod-in-reference-frame-of-COM}$$

*****************************************

So that's all with the terms. Now about your approach, I would say the part about angular momentum conservation is wrong. It's not because you don't understand; it's because the mistake is a very very subtle point about angular momentum that many introductory textbooks ignore. Caution must be taken when dealing with angular momentum. Going into detail about this takes time, so I'll leave it aside (if you're interested, you can search my replies to topics by PhMichael; I remember that I once discussed about it somewhere). Usually there are 2 cases that we can be confident when applying the angular momentum conservation:
_ When the point A (the point about which we calculate angular momentum) is a fixed point, provided that the total torque about A is zero.
_ When the point A is COM, provided that the total torque about A is zero.

In this problem, A (the point of contact) is neither a fixed point nor COM, so we should avoid any equation related to angular momentum about A. Instead we can consider COM (so that we can write down equations without reservation!). Let X denote the impulse by the ground, then it obviously points upwards.
_ From Newton's 2nd law: $$X = m(V - (-v_0))$$
_ Since X also creates a change in angular momentum about COM: $$Xlcos\alpha = \frac{1}{12}ml^2\omega$$
(this comes from this popular equation: $$torque = FL = I\gamma = Id\omega/dt$$
thus: $$I\int d\omega = L\int Fdt = LX$$)

From the 2 equations, we obtain the relation between V and w. Plus the energy conservation equation, the problem is solved

Haha, sorry for the long-winded reply

 

[zorro]

Thanks a lot for your detailed reply.
The only thing wrong in my approach was that I considered angular momentum about point A.
Now if I consider Angular Mometum about the COM,

mv_olcosϑ/2 = ml²w/12 + mVlcosϑ/2
from here,
cosϑ(v_o - V)=wl/6

I thought of a new approach from here
What if we apply Newton's Law of restitution with e=1
i.e. ev_o= -V
I got w=12v_ocosϑ/l

Since X also creates a change in angular momentum about COM: $$Xlcos\alpha = \frac{1}{12}ml^2\omega$$

I guess u meant Xlcosα/2=ml²w/12
Using your idea, I got
http://latex.codecogs.com/gif.latex...}&space;+&space;v_{o}\right)\frac{12}{l}

 

[hikaru1221]

I guess u meant Xlcosα/2=ml2w/12

Yep, I forgot the number 2
I'm not familiar with Newton's law of restitution, but one question: Is the law proved to cases where rotational motion is taken into account?

And I cannot read our final result (check your Latex code ). Anyway, one problem:

Now if I consider Angular Mometum about the COM,

mvolcosϑ/2 = ml2w/12 + mVlcosϑ/2

Is the torque about COM equal to zero?

 

[aim1732]

mvolcosϑ/2 = ml2w/12 + mVlcosϑ/2

I think it would be right and much easier to conserve angular momentum about the point of contact with the floor as the unknown impulse passes through that point and generates no torque about it.In that regard you have already done that although you think otherwise-a translating body can not have angular momentum about it's centre of mass and the first term should have been zero. Anyways it would be wrong to conserve L about com because of the impulsive torque of the reaction of the ground.
To hikaru1221:

Is the law proved to cases where rotational motion is taken into account?

The law is always valid but in rotational motion of rigid bodies different points have different velocities and hence it is applied only about points. In this particular case the velocity along the normal of the point of contact of the rod is reversed while it acquires some general velocity along the tangent(the latter is not Newton's law of restitution).

One thing the op can do is:
Combine rotation and translation of point of contact along the normal.This will give:
w(l/2)cosϑ - vc = v1
where w and vc are the angular velocity of the system and velocity of centre of mass after the impact.

 

[zorro]

http://latex.codecogs.com/gif.latex...qrt{12}}&space;+v_{o}\right)\frac{12}{l}

Torque about COM will be 0 if we consider the wall and the rod as our system.
If we take only rod into consideration, torque will not be 0.

Combine rotation and translation of point of contact along the normal.This will give:
w(l/2)cosϑ - vc = v1
where w and vc are the angular velocity of the system and velocity of centre of mass after the impact.

Newton's Law of restitution is valid if we take the rotation and translation about Centre of mass only. Moreover, cosϑ term won't be there. V_c will be positive (if it is negative, w(l/2) should also be negative)

V + wl/2 = v_o

But substituting this in the equation
cosϑ(v_o - V)=wl/6
we get a value of cosϑ as V vanishes

I think the energy conservation is the best approach!

 

[hikaru1221]

http://latex.codecogs.com/gif.latex...qrt{12}}&space;+v_{o}\right)\frac{12}{l}

Eek, again, I cannot see it

Torque about COM will be 0 if we consider the wall and the rod as our system.
If we take only rod into consideration, torque will not be 0.

Then you have to take into account the angular momentum of the wall (or earth). Not easy though!
(notice that the velocity of the Earth due to the impact seems to be negligible, but its mass is huge and so is its moment of inertia. In the end, its angular momentum may be significant)

EDIT: By the way, COM of the system is not COM of the rod.

Newton's Law of restitution is valid if we take the rotation and translation about Centre of mass only. Moreover, cosϑ term won't be there. V_c will be positive (if it is negative, w(l/2) should also be negative)

V + wl/2 = v_o

But substituting this in the equation
cosϑ(v_o - V)=wl/6
we get a value of cosϑ as V vanishes

I think the energy conservation is the best approach!

I'm not sure about the Newton's law of restitution. Let's see if the 2 answers match each other.

 

[aim1732]

Newton's Law of restitution is valid if we take the rotation and translation about Centre of mass only.

What is this supposed to mean?

Moreover, cosϑ term won't be there

If the body were purely rotating about the centre of mass then it's velocity would have been wl/2 in direction perpendicular to the rod, hence along the normal to the ground there is a cos component.And since the torque is clockwise so will be w and the velocity due to rotation will be opposite to that due to translation.

Torque about COM will be 0 if we consider the wall and the rod as our system.
If we take only rod into consideration, torque will not be 0.

I don't know if that's right.It gives the same equation that you will get by conserving the L of the rod only, unless you plan on quantifying the angular momentum of the wall.

 

[zorro]

Eek, again, I cannot see it

See my attachment

What is this supposed to mean?

If you want to use Newton's Law of restitution for a body that translates as well as rotates, it should be applied about the centre of mass as different points of the rod have different velocities.

If the body were purely rotating about the centre of mass then it's velocity would have been wl/2 in direction perpendicular to the rod, hence along the normal to the ground there is a cos component.And since the torque is clockwise so will be w and the velocity due to rotation will be opposite to that due to translation.

There is no change in angular momentum of the rod along x-direction (surface is frictionless).
So I think the the velocity vector will be normal to the ground.

 

[zorro]

@Hikaru
Open the attachment in a new tab.

 

[hikaru1221]

Eek, you should recheck the formula. Why is there omega in the final answer?
Anyway:
_ From 2 equations of post #6:
$$X = m(V+v_0)$$
$$Xlcos\theta/2 = ml^2w/12$$
We have: $$wl = 6cos\theta(V+v_0)$$
_ From energy conservation law: $$0.5mv_0^2=0.5mV^2+0.5\times m(wl)^2/12$$
From those 2 equations, you get a quadratic equation. Solve it for V. One solution is $$-v_0$$, which is eliminated due to that if $$V=-v_0$$ , there would be no change in COM's velocity, which means there would be no impact / collision at all (X=0!), and that's impossible.

 

[zorro]

OHH! I forgot that we had to find an expression for Angular velocity and not V.
I got this answer
w= v_o(1-3cos²ϑ)/(1+3cos²ϑ)
Thanks for spending your precious time

 

[aim1732]

If you want to use Newton's Law of restitution for a body that translates as well as rotates, it should be applied about the centre of mass as different points of the rod have different velocities.

I think you seriously need to consider the validity of this statement. From what I have known the law is applied on the points on rigid extended bodies that come in contact during the course of the collision.

There is no change in angular momentum of the rod along x-direction (surface is frictionless).
So I think the the velocity vector will be normal to the ground.points of contact of colliding extended rigid bodies.

There is NO angular momentum along the x-direction[it is not like linear momentum].
And as for velocity of the end of the rod in contact with the ground it is not conclusively along the normal to the floor-you have not proved it. It may be true but physics is not built on grand proclamations.