A mass collides with a stick and they stick together....

[Den tredje]

Homework Statement

Exam

Relevant Equations

1/2Iw^2 = mgh
(1/2(1/12Ml^2+(1/2)^2Mw^2)/mg = h = 0.157

(Note: the stick weight is 0.4kg not 1.0kg)

Im not a native English speaker. So please excuse my bad english.

I had this on my exam, and Ím really unsure what the answer to this is. Ím suppose to calculate the hight and angle of the swing Of this stick after a “stuck together” collition. But I’m really unsure if I should calculate a new masscenter or is it enough to just calculate the moment of inertia?

It confuses me. Since I think if I tilt the stick 30degrees before the crash, the gravity of the tip is the same? So it will stand still if I tilt it 30degrees? It feels unlogical, but I am not able to make a proper argument to explain my feeling. I would be great with either an answer or explanation what would be right/wrong to think...

 

[etotheipi]

Energy is not conserved during the collision (it's inelastic), but angular momentum about the axle is (why?). After the two have stuck together again, the energy of the ball/stick system from then on is conserved (why?).

Can you use that to make a start?

 

[haruspex]

So it will stand still if I tilt it 30degrees?

If you just tilted the stick, no attached mass, yes. The axle is at the stick's mass centre, so tilting the stick does not change its gravitational potential energy. The arms extending each side exert equal and opposite torques.

 

[Den tredje]

If you just tilted the stick, no attached mass, yes. The axle is at the stick's mass centre, so tilting the stick does not change its gravitational potential energy. The arms extending each side exert equal and opposite torques.

Thank you, this confuses me a lot considering solving this. Since if you attatch a ball to the tip, its not equal anymore. Which makes me think I can consider the stick as a weightless rope, with the ball in the end of it? Because the stick cancel itself out. Like if I was to give the stick a spin in around its axis, it will spin indefinitely. I am sure it will spin indefinitely without friction.

With a ball attached to the end of it. I am not able to propperly image how the movement of the stick is. Since the center of mass is not on the middle of the axis. and its not on tip of the stick either, like it would be in a weightless rope sort of situation. I thought I could consider the center of mass as my new ball. And consider there was a weightless rope between my axis to my center of mass as a very short rope, and then just scale it properly afterwards.

But I am not sure, if I can consider it as a weightless rope with a ball swinging back and forth.

Energy is not conserved during the collision (it's inelastic), but angular momentum about the axle is (why?). After the two have stuck together again, the energy of the ball/stick system from then on is conserved (why?).

Can you use that to make a start?

Hmmm... english...
So with energy conserved you are thinking of what I’m saying above. That its conserved because it starts swinging back and forth? Which I’m not sure is correct In this chase...

Are you testing me that I know if you changes the length of the stick, and or mass, etc it will have the same factors. Like with I shorter radius the speed increases etc to uphold the momentum? And more mass wil just slow the system down Etc?

Energy is conserved because it will have kinetic energy in the bottom and potential energy at the top of the swing. I’m not sure my English is good enough to explain this properly in English. Though I don’t think I have a problem with this part?

Unless you are hinting that I should consider the crash first, then calculate the energy to get the hight? That’s... Hmm... That’s not what I did no. ^^”

 

[haruspex]

Since the center of mass is not on the middle of the axis. and its not on tip of the stick either

In my experience, finding the centre of mass of a composite system is rarely worthwhile. It is simpler to consider the contribution of each part (whether to torque or moment of inertia etc.) separately and add them up.

it would be in a weightless rope sort of situation.

The weight of the stick is irrelevant, yes, but in the collision its mass is important.

Unless you are hinting that I should consider the crash first, then calculate the energy to get the hight?

Yes, that is what you must do.

 

[etotheipi]

Unless you are hinting that I should consider the crash first, then calculate the energy to get the hight? That’s... Hmm... That’s not what I did no. ^^”

The reason is that in the initial collision some energy is dissipated as heat and becomes thermal energy of the ball/stick system. The total energy is still conserved if you look hard enough, but kinetic energy is not and this approach isn't feasible. But the angular momentum of the ball-stick system about the axle is, since the reaction force at the axle produces no torque about the axle and the reaction forces between the ball and the stick are internal.

 

[Den tredje]

The reason is that in the initial collision some energy is dissipated as heat and becomes thermal energy of the ball/stick system. The total energy is still conserved if you look hard enough, but kinetic energy is not and this approach isn't feasible. But the angular momentum of the ball-stick system about the axle is, since the reaction force at the axle produces no torque about the axle and the reaction forces between the ball and the stick are internal.

Yes, I agree. I’m just confused It being a stick. I think. Thank you.

In my experience, finding the centre of mass of a composite system is rarely worthwhile. It is simpler to consider the contribution of each part (whether to torque or moment of inertia etc.) separately and add them up.

The weight of the stick is irrelevant, yes, but in the collision its mass is important.

Yes, that is what you must do.

Composite system means a system with several parts to it right?

ok, so what you are saying is that I should say moment of the ball is = moment to the entire system? and ignore the stick when doing the energy calculations To get the height.

About your experience. Could you give an example? Thank you.

 

[haruspex]

I should say moment of the ball is = moment to the entire system?

No, the stick has mass and length, so it has moment of inertia. I am saying that (for just after the collision) you can compute the moments of inertia of ball and stick about the axle separately and add them to get the moment of inertia of the combination.

ignore the stick when doing the energy calculations To get the height.

Not entirely. You can ignore its weight, so you can ignore its gain in gravitational potential energy. But it will lose rotational KE, which you cannot ignore.

 

[Den tredje]

No, the stick has mass and length, so it has moment of inertia. I am saying that (for just after the collision) you can compute the moments of inertia of ball and stick about the axle separately and add them to get the moment of inertia of the combination.

Not entirely. You can ignore its weight, so you can ignore its gain in gravitational potential energy. But it will lose rotational KE, which you cannot ignore.

ah, ignore it in the context of how the calculation goes. Yeah...

It crashes, then it flies. Thanks! ;) Ill see to post my final calculation When I get there. Thanks for great answers.

 

[etotheipi]

ah, ignore it in the context of how the calculation goes. Yeah...

No he's saying you can't completely ignore it for the calculation. For the change in GPE, fine, you only need consider the height increase of the ball. But in order to calculate the initial KE $= \frac{1}{2}I\omega^2$, the whole configuration is initially rotating so you must also add the moment of inertia of the stick.

 

[Den tredje]

No he's saying you can't completely ignore it for the calculation. For the change in GPE, fine, you only need consider the height increase of the ball. But in order to calculate the initial KE $= \frac{1}{2}I\omega^2$, the whole configuration is initially rotating so you must also add the moment of inertia of the stick.

That is what I felt I said. Sorry for my English.