Pointmass colliding with windmill

In summary, the conversation discusses a problem involving a toy windmill and a ball of mass m. The windmill consists of four uniform rods arranged at right angles, and the moment of inertia about the axle is ##I = \frac{4ml^2}{3}##. The ball is dropped from a height h and makes an elastic collision with the windmill. The problem asks for the angular velocity of the windmill after the collision and the velocity of the ball after the collision. The equations used to solve the problem are potential energy, kinetic energy, momentum, and angular momentum. Some confusion arises regarding the conservation of angular momentum, but it is clarified that it is conserved about the axle. The summary concludes by mentioning that the
  • #1
Alettix
177
11

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! :)
 

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  • #2
Alettix said:
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.
 
  • #3
kuruman said:
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?
 
  • #4
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.
 
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  • #5
Alettix said:
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.
Alettix said:
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)##
 
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  • #6
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.
 
  • #7
PeroK said:
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!
 
  • #8
kuruman said:
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.
 
  • #9
Alettix said:
unknown force at the axel
Better is "unknown impulse". The force is variable during the brief impact, and anyway unknowable.
Alettix said:
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?
 
  • #10
haruspex said:
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.
 
  • #11
Alettix said:
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.
 
  • #12
haruspex said:
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!
 

Related to Pointmass colliding with windmill

1. How does a point mass colliding with a windmill affect its rotation?

When a point mass collides with a windmill, it transfers its momentum to the blades of the windmill. This sudden increase in momentum causes the windmill to rotate faster.

2. What factors determine the impact force of a point mass colliding with a windmill?

The impact force of a point mass colliding with a windmill is determined by the mass and velocity of the point mass, as well as the shape and construction of the windmill.

3. Can a windmill be damaged by a point mass collision?

Yes, a windmill can be damaged by a point mass collision. Depending on the speed and mass of the point mass, the blades of the windmill may bend or break upon impact.

4. What happens to the energy of a point mass colliding with a windmill?

When a point mass collides with a windmill, its kinetic energy is transferred to the windmill's blades. This energy is then converted into rotational energy, causing the windmill to spin faster.

5. How do collisions between point masses and windmills affect wind energy production?

Collisions between point masses and windmills can potentially increase the efficiency of wind energy production. The added momentum from the collision can increase the rotational speed of the windmill, resulting in more energy being generated.

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