Halp a kinetically impared anon - physics salvation thread.

>>7978066
>kinetics

Also, what the fuck do you mean by visible mass at the point of impact? What the actual fuck?

>>7978089
I think he means how much momentum is transferred to the struck object. Not sure how to work. Really we would need to know the properties of this "immovable object" though. Hopefully it's rubber or something.

>>7978093
If the object is by definition immovable, then no momentum can be transferred to it.

>>7978102
So I guess I mean energy transfer in the form of a wave through it?

>>7978105
The only energy transfer that will happen is inside the moving object, which will lose kinetic energy and deform.

>>7978089
I mean if one drops a block as such it will ding that edge and get some torque and lose some amount of it's linear momentum.
So the force reflected back into the block from the impact event can't be the mass of the entire block times the velocity at the point of impact since that would cancel out the downward motion entirely.
(or momentarily bringing it to a full halt if falling under gravity)

So since the normal force of the impact do not stop the block it can only be a fraction of the blocks mass that contributes to the impact right?
That is what I mean by 'visible mass' I don't know the correct terminology for this, hence why I'm asking.

>>7978119
I really don't know what you're asking. The second block is immovable, it will reflect all forces and the reflected force will result in the deformation of the moving block and the loss of velocity.

>>7978193
I'm asking what the magnitude of that impacting force is.
Like if I stuck a scale at the point of impact, what would it tell me about the impulse?

The impact point in the example comes into contact with the imovable object at a velocity of 10 meters per second.
so the velocity both the surfaces comes into contact is ofcourse exactly that. but what happens now?

According to my understanding energy is now sent into the imovable object and reflected back into the moving object along the direction of the normal of the impact.
This now creates both rotational and translational forces generating a counterclockwise torque around the lever arm between center of mass and impact point
and also translating the center-mass along the same vector as the normal of the impact, with a magnitude equal to the magnitude of the impact.

But if you drop or throw a block against a edge in this manner you can see that the center of mass of that block continous moving at largely unchanged linear velocity with the object gainging a lot of angular momentum.
So since the center of mass wont come to a full halt in a impact like this, that should mean only a fraction of the mass of the moving object counts towards this off-center collision event?
Hence my question, what fraction of the total mass is 'visible' behind the point comming into contact with the imovable object?

1) the entirety of the falling block mass

2) none of the immovable block mass

your system is equivalent to a force F applied to the middle of the contact surface between the two objects, with a magnitude of gravity directed to the top

>>7978248
>your system is equivalent to a force F applied to the middle of the contact surface between the two objects, with a magnitude of gravity directed to the top

But if that is true why doesn't a falling box dinging into a edge come to a halt during impact?
like if the block is falling at the speed of sound and slams into the corner woulden't it instantly gain a linear velocity of zero if what you say is correct and the collision is inelastic?
or fly of spining like crazy in the exact direction it came from if the collision is elastic.

File: offcenterImpact.jpg (87KB, 843x848px) Image search: [Google]

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>>7978297
>>7978248

This perhaps serves to clarify my issue with this.

>>7978297
>>7978330

1) even if it's perfectly inelastic, it'll still spin

2) the centermass won't receive the same impulse in your picture

3) you'll have to take into account the entire mass of the falling object but it'll be multiplied by a constant depending of the distance between the center of mass of the falling object and the poibt of impact

>>7978358
>3) you'll have to take into account the entire mass of the falling object but it'll be multiplied by a constant depending of the distance between the center of mass of the falling object and the poibt of impact

How do I go about finding the equation of that constant, what is it called?

bumptasticum

>>7978377
Anyone?

>>7978330
So some of the kinetic energy from the falling block is converted into rotational energy. None is created.

In other words, if the block gains rotational energy, it will lose momentum.

The loss of momentum is proportional to the "felt impact mass", which you are interested in.
If the block loses 30% of its momentum, its felt impact mass is also only 30%.
And if it loses 100%, well... go figure.

The amount of gained rotational energy can be calculated quite easily i guess. Obviously, its point of impact has to rotate as fast as the block was falling. Or something like that.

File: 1456331962404.png (1MB, 852x854px) Image search: [Google]

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>>7978297
>But if that is true why doesn't a falling box dinging into a edge come to a halt during impact?
>what is newton's second law

>>7979510
But Newtons second law is essentially F = ma, how does that explain anything? I don't know what the effective mass behind the impact is and I don't know the force I only know the impact velocity.

As >>7978358 says it must be a multiplier of the mass based on the distance from the center mass, but I can't find or figure out the equation of that number.
I imagine it must involve placing an imaginary pivot at some point and measure how hard the mass would torque down on the contact point..

Yet here you state Newton's 2nd law while giving me the fish having a hamburger stare like I am missing something completely obv... Am I missing something obvious?

Ok so you're looking for m in F=ma
well just look at the amount of mass that actually hits at the time of impact.
Draw a line upwards from the edge of the immovable object. Everything on the lines right side counts towards that m.

If the second block doesn't move - there's no energy transfer to it. The first block retains all original energy.
While falling straight it has pure kinetic energy, [math]E_{total}=T=\frac{p^2}{2m}[/math].
When it strikes the block it transfers some kinetic energy into rotational energy:
[math]E_{total}=T+R=\frac{p'^2}{2m}+\frac{L^2}{2I}[/math]
By considering conservation of energy for the moment immediately after the collision, you can work out the amount of energy that was transferred from kinetic to rotational, and this should be the amount that was available to the second block.

Source: Physics

File: trainCrush.jpg (77KB, 1214x656px) Image search: [Google]

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>>7980322
That can't be right, pic related is why.

File: intuition.jpg (77KB, 1130x656px) Image search: [Google]

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Is it legit to think of this situation like pic related? like what force would be needed to stop the mass from moving had it been supported at a virtual pivot unit distance away?

My intuition tells me the solution must look something like this but surely someone here must know?

>>7980915
To stop it from turning, you need to create a torque that equals the torque from the mass, in the opposite direction.

Is the virtual pivot the point the COM is coming from?

>>7980971
I'm trying to figure out what force a free-falling object generates on off-center impact. This is my attempt at answering my own unresolved Op-question.

Center mass in this example is located at the red circle, in the center of the locomotives mass.

File: scan.jpg (103KB, 571x926px) Image search: [Google]

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>>7980150
>Am I missing something obvious?
you're assuming that the force on the impact/contact point cannot be equal to the object mass times the gravitational constant
your logic behind this is that if the force would be equal to m*g, the object would be 'instantly stopped', as evident by your post >>7978297 ("instantly gain a linear velocity of zero")

this is wrong, because forces don't simply cause velocities to vanish
F=ma, a force acting on mass causes acceleration(/deceleration), it does not cause a mass to become stationary
you're applying statics to an inherently dynamic problem and make skewed assumptions in the process
I'm not saying that the force IS equal to the object mass times the gravitational constant
but you're reasoning as to why this is the case is wrong

the force acting off-center of the already moving object causes both a linear deceleration and an angular acceleration
in order to get an actual value for the force you need to know the objects rotational inertia first
read through the pic and try to understand what's being done step by step

dphi/dt^2 = angular acceleration
Js = rotational inertia about the COG
Ja = rotational inertia about the point/axis A (through the parallel axis theorem)
F1 = the force you're looking for

as you can see by the results the force is dependent on
-mass
-overall rotational inertia
-distance between center of gravity and contact point
-gravitational constant

Thank you for taking your time to explain anon, saved a printscreen of this post, gonna work trough your answer soon as I wake up tomorrow.

I need all this for a basic rigid body physics simulation I'm writing, perhaps my concept of impulses and forces are getting mixed together and need some sorting out.
Also googled this 'parallel axis theorem', a quick read suggests this 'second moment of area' def is a concept I've been grasping after.