Pointmass colliding with windmill

[Alettix]

Homework Statement

A toy windmill consists of four thin uniform rods of mass m and length l arranged at rigth angles, in a vertical plane, around a thin fixed horizontal axel about which they can turn freely. The moment of intertia about the axel is $I = \frac{4ml^2}{3}$.
Initially the windmill is stationary. A small ball of mass m is dropped from a height h above the end of a horizontal rod, with which it makes an elastic collision. What is the angular velocity of the windmill after the collision and with what velocity does the ball rebound?

Homework Equations

Potential energy: $E_p = mg\Delta h$
Kinetic energy: $E_k = \frac{mv^2}{2} + \frac{I\omega^2}{2}$
Momentum: $p=mv$
Angular momentum: $L=I\omega$

The Attempt at a Solution

The speed the speed the ball hits the windmill arm with is:
$v = \sqrt{2gh}$
Let $u$ be the velocity it rebounds with, and $\omega$ the angular velocity of the windmill.

Because the windmill cannot move translationally, translational momentum cannot be conserved. This means that an external force has to act at the axel to keep it back from accelerating downwards. Because and external force acts on the system, angular momentum is not conserved either (or is it? I am unsure about this, maybe it is conserved about the centre of the windmill, but I need some guidance here).

When the ball collides with the windmill it acts with an impulse, giving it an angular momentum:
$L_w = l \Delta p = l m (u-v)$ (where $u$ is a negative quantity and hence the windmill will rotate downwards). This leads to:
$\omega = \frac{3(u-v)}{4l}$

If we now assume that the energy is conserved (elastic = totally elastic?):
$\frac{mv^2}{2} = \frac{mu^2}{2}+\frac{I\omega^2}{2}$
Inserting our previous expression for $\omega$ this equation gives:
$u = \frac{9-\sqrt{109}}{14} v$ and hence: $\omega \approx \frac{0.827v}{l}$.

The problem is that this is all wrong. The answer is supposed to be $u = v/7$ and $\omega = \frac{6v}{7l}$. I am unsure about where my solution goes wrong. Is the assumption of energyconservation faulty? Is MoI conserved about the axel? But how can MoI be conserved about some points but not others?

Thank you for any help! :)

 

[kuruman]

Because and external force acts on the system, angular momentum is not conserved either (or is it?

Angular momentum is conserved in a given direction if no external torques act in that direction. Here you can pretend that it is conserved instantaneously while the collision takes place.

 

[Alettix]

Angular momentum is conserved in a given direction if no external torques act in that direction. Here you can pretend that it is conserved instantaneously while the collision takes place.

Can I pretend it is or is it conserved? Is it conserved about all points or just the axel where the external torque is zero?

 

[kuruman]

The external torque on the falling mass due to gravity is not zero about the axle. It is mgl. However, you can pretend that the collision happens so fast that the angular momentum of the ball just before the collision is equal to the angular momentum of ball plus windmill just after the collision. You are interested in angular momentum about the axle because in $I \omega$ the moment of inertia $I$ is about the axle, not some other point.

 

[PeroK]

When the ball collides with the windmill it acts with an impulse, giving it an angular momentum:
Lw=lΔp=lm(u−v) L_w = l \Delta p = l m (u-v) (where uu is a negative quantity and hence the windmill will rotate downwards). This leads to:
ω=3(u−v)4l

This is correct.

If we now assume that the energy is conserved (elastic = totally elastic?):
mv22=mu22+Iω22

And this. So, perhaps it's just your algebra at fault.

PS That said, you shouldn't need to solve a quadratic, as:

$v^2 - u^2 = (v-u)(v+u)$

 

[PeroK]

PS Gravity is not relevant to the collision here, as these problems tacitly assume an instantaneous collision. Although, IMHO, they should always say so explicitly.

 

[Alettix]

This is correct.And this. So, perhaps it's just your algebra at fault.

PS That said, you shouldn't need to solve a quadratic, as:

$v^2 - u^2 = (v-u)(v+u)$

Alrigth, I will check through my algebra!

 

[Alettix]

The external torque on the falling mass due to gravity is not zero about the axle. It is mgl. However, you can pretend that the collision happens so fast that the angular momentum of the ball just before the collision is equal to the angular momentum of ball plus windmill just after the collision. You are interested in angular momentum about the axle because in $I \omega$ the moment of inertia $I$ is about the axle, not some other point.

Oh, I understand thank you!
Just out of curiosity: if we disregard gravity completely and only care about the unknown force at the axel (say the windmill is laying horizontal on a table and a rolling ball hits it): would angular momentum be conserved about any point or just the axel where the torque from the external force is zero? The perpendicular axis theorem could be used to find the MoI of the windmill about any point.

 

[haruspex]

unknown force at the axel

Better is "unknown impulse". The force is variable during the brief impact, and anyway unknowable.

would angular momentum be conserved about any point or just the axel where the torque from the external force is zero?

What do you think? If there is an unknown external impulse and its line of action is not through the reference axis, would it alter the angular momentum about that axis?

 

[Alettix]

What do you think? If there is an unknown external impulse and its line of action is not through the reference axis, would it alter the angular momentum about that axis?

Yes it would. So angular momentum is assumed to be conserved about the axel only and I must have done a misstake in my algebra when calculating the velocity u.

 

[haruspex]

Yes it would. So angular momentum is assumed to be conserved about the axel only and I must have done a misstake in my algebra when calculating the velocity u.

Yes, the equations you posted do lead to the rfght answer (but with the sign of u reversed). If still stuck, please post your working.

 

[Alettix]

Yes, the equations you posted do lead to the rfght answer (but with the sign of u reversed). If still stuck, please post your working.

Fixed it, Thanks a lot for the help everybody!