Momentum -- pure rolling and trajectory physics problem

In summary: vi is the velocity at the top of the wall before the collision vf is the velocity after the collision wi is the angular velocity of the ball right after the collision gμt is the angular momentum of the ball right after the collision t is the time it takes for the ball to acquire this angular momentum
  • #36
oh then vi = wir? why are you asking that
 
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  • #37
vishnu 73 said:
oh then vi = wir? why are you asking that
So combine that with your first two equations in post #34.
 
  • #38
no but aren't you assuming it transits immediately to pure rolling ? as the impulses we denoted in post #34 are only when the ball hits the ground which is a very short time
 
  • #39
vishnu 73 said:
no but aren't you assuming it transits immediately to pure rolling ? as the impulses we denoted in post #34 are only when the ball hits the ground which is a very short time
I'm not assuming it transits immediately to rolling. I am trying to find out if that is possible.
The brevity of the impact is irrelevant. The impulse is ∫F.dt. It can be a very strong force for a very short time.
 
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  • #40
ok i kind of get what you are saying now so i did this:
no more vf and wf and t

mvx = Jx = mwir
mwir = Jx r2m/I
solving for Jx i got
Jx = 2mvx/7
then what the other equation we have is
jx ≤ μJv am i supposed to use this
is what i did correct
 
  • #41
vishnu 73 said:
ok i kind of get what you are saying now so i did this:
no more vf and wf and t

mvx = Jx = mwir
mwir = Jx r2m/I
solving for Jx i got
Jx = 2mvx/7
then what the other equation we have is
jx ≤ μJv am i supposed to use this
is what i did correct
Yes, that all looks good. It should give you the relationship between μ, vx and vy which allows instantaneous transition to rolling.

(I was thrown at first by a typo, = for -, in your first equation above.)
 
  • #42
oh you sorry about the typo
i don't quite get what you are saying after that we know vx we know μ how are we even supposed to find time from that equations
 
  • #43
vishnu 73 said:
how are we even supposed to find time from that equations
If we can show the transition is instantaneous upon landing, how long does the transition take?
 
  • #44
isn't that instantaneous?
 
  • #45
vishnu 73 said:
isn't that instantaneous?
Quite so, but you asked how we would find time.
 
  • #46
how do you derive time from these equations the only thing i see the impulse term having time as force * time but what is force help me please!
 
  • #47
vishnu 73 said:
how do you derive time from these equations the only thing i see the impulse term having time as force * time but what is force help me please!
An impact like this takes an unknowable time and generates unknowable forces. All we can know is the impulse, mΔv=∫F.dt. The elapsed time is assumed arbitrarily short, effectively zero. That is what instantaneous means. If the frictional impulse during the impact is sufficient to lead to rolling contact then the answer to the question is zero.
So, is it sufficient or not? To find out, obtain the relationship between vx, vy and μ.
 
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  • #48
but the answer given in the brilliant org was 1/15 sec but i think i trust you more
so you are saying that if Jx = Jvμ then there is instantaneous transition to pure rolling
so since Jx here is 2mvx /7 ≤μmvy hence there is immediate transition to pure rolling
is my understanding correct
 
  • #49
vishnu 73 said:
but the answer given in the brilliant org was 1/15 sec but i think i trust you more
so you are saying that if Jx = Jvμ then there is instantaneous transition to pure rolling
so since Jx here is 2mvx /7 ≤μmvy hence there is immediate transition to pure rolling
is my understanding correct
Yes, that's it.
 
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  • #50
so before each problem i should check whether the impulse is enough to cause pure rolling instantly?
 
  • #51
vishnu 73 said:
so before each problem i should check whether the impulse is enough to cause pure rolling instantly?
Yes.
 
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  • #52
ok i have a few more questions not related to this thread but going to post them as they are just trivial doubts

ok so if ball is rolling down a hill we say it has rotational and translational kinetic energy but when a disk is spinning about its axis we only say it has rotational kinetic energy why is that so?
 
  • #53
vishnu 73 said:
ok i have a few more questions not related to this thread but going to post them as they are just trivial doubts

ok so if ball is rolling down a hill we say it has rotational and translational kinetic energy but when a disk is spinning about its axis we only say it has rotational kinetic energy why is that so?
It is common to regard the velocity of the mass centre as constituting translational KE, with any other motional energy considered rotational. But there other ways of considering it. A rolling disk or ball is, at each instant, rotating about its point of contact with the ground. Thus, it could all be considered rotational energy.
 
  • #54
so it is just that about the instantaneous center i only regard the rotational energy?(Adjusting MOI)

then what is energy of the Earth around the sun ignoring Earth's spin on its own axis
is it GPE + translational KE + rotational KE about the sun?
 
  • #55
vishnu 73 said:
so it is just that about the instantaneous center i only regard the rotational energy?
No, as I wrote, it can be viewed either way. The total KE is the same.
vishnu 73 said:
then what is energy of the Earth around the sun ignoring Earth's spin on its own axis
That's different. There is no rotational KE here since no rigid body is rotating.
 
  • #56
wait what is rigid body i mean is there a formal definition for a rigid body?
are you saying it does not have rotational KE because the sun and the Earth are not connected?
 
  • #57
vishnu 73 said:
wait what is rigid body i mean is there a formal definition for a rigid body?
are you saying it does not have rotational KE because the sun and the Earth are not connected?
At root, the KE of a system is simply ∑½mivi2, where the sum is taken over infinitesimal mass elements. Each such element is considered so small as to have no significant rotational KE.
But that is inconvenient when analysing systems of countless such elements. For a rigid body, i.e. one in which the pairwise distances between every pair of mass elements in it are conserved, we can represent the entire motion as the sum of a linear motion of its mass centre plus a rotational motion about some axis through that centre. For such a representation, it can be shown that ∑½mivi2=½(∑mi)vcom2+½Ixω2, where ω is the rate of rotation and Ix is a constant describing the distribution of mass within the body about the given axis. Specifically, Ix=∑mixi2, where xi is the distance of element i from the axis.

In the case of a planet orbiting its star at distance r, not rotating on an axis of its own (i.e. one year=1 day, not gravitationally locked), each little mass element is describing a circle at the same rate and the same radius, just not all around the same centre. So if the linear speed of the planet is v then the KE of the planet is ∑½mivcom2=½Mvcom2.
 
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  • #58
oh now i get helped when you pointed out the part with equations thanks a lot was really confused by this over the past few days thanks once again
well here is another doubt of mine not exactly a doubt per say but a little thought question by myself
so say a ball is rolling on table (pure rolling) then it enters a paper on the table and continues to pure roll
then we suddenly pull the paper opposite to direction of the motion of the ball
can you please explain what exactly is happening to the ball when we suddenly pull the paper?
thanks your help is much appreciated
 
  • #59
vishnu 73 said:
say a ball is rolling on table (pure rolling) then it enters a paper on the table and continues to pure roll
then we suddenly pull the paper opposite to direction of the motion of the ball
can you please explain what exactly is happening to the ball when we suddenly pull the paper?
So, the ball is rolling to the right, say, and we yank the paper suddenly to the left.
Kinetic friction will act to the left on the ball, slowing it linearly but accelerating its rotation. As long as we pull the paper too fast for rolling to be re-established, those changes will continue, perhaps with the ball starting to move to the left.
Is that your question?
 
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  • #60
ok say we pull it slowly then what happens
 
  • #61
vishnu 73 said:
ok say we pull it slowly then what happens
In that case we do not immediately know what the frictional force is. Instead, we know the relationship between the frictional force and the ball's linear acceleration, the relationship between the frictional force snd the angular acceleration, and the relationship between those two accelerations given that it continues to roll without slipping. Three equations, three unknowns.
 
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  • #62
ok thanks helped a lot that was all the doubts i had thanks once again
 
  • #63
If both spheres have the same radius, then I dont' see how there can be any transfer of momentum to cause 'oblique projectile motion'. It seems to me the only way this can happen is if the 1m mass actually has a larger radius than the 2m mass such that the line of collision is also 'oblique' allowing for a vertical component in the momentum transfer. The 1m mass 'jumps' when it hits the 2m mass.
 
  • #64
neilparker62 said:
If both spheres have the same radius, then I dont' see how there can be any transfer of momentum to cause 'oblique projectile motion'. It seems to me the only way this can happen is if the 1m mass actually has a larger radius than the 2m mass such that the line of collision is also 'oblique' allowing for a vertical component in the momentum transfer. The 1m mass 'jumps' when it hits the 2m mass.
There are several impossibilities in the specification of this question. I would not waste any time on it.
 
<h2>1. What is momentum and how is it related to pure rolling and trajectory physics?</h2><p>Momentum is a measure of an object's motion, determined by its mass and velocity. In pure rolling, an object is rolling without slipping, meaning that its point of contact with the ground is stationary. In trajectory physics, momentum is important because it determines the path an object will take when it is launched into the air.</p><h2>2. How is momentum conserved in pure rolling?</h2><p>In pure rolling, momentum is conserved because there is no external force acting on the object. This means that the total momentum of the object before and after rolling remains the same. This is because the object's rotational and linear velocities are directly related to each other, and any changes in one will be offset by changes in the other to maintain the same total momentum.</p><h2>3. What factors affect the trajectory of an object in motion?</h2><p>The trajectory of an object is affected by its initial velocity, the angle at which it is launched, and the force of gravity. Air resistance can also play a role in altering the trajectory of an object.</p><h2>4. How can we calculate the momentum of an object in pure rolling?</h2><p>The momentum of an object in pure rolling can be calculated by multiplying its mass by its linear velocity. This is because the object's rotational and linear velocities are directly related, and the total momentum is the sum of both types of velocity.</p><h2>5. Can momentum be transferred between objects in pure rolling?</h2><p>Yes, momentum can be transferred between objects in pure rolling if there is an external force acting on the objects. For example, if two objects collide while rolling, the momentum of one object will be transferred to the other object. However, if there are no external forces, the total momentum of the system will remain constant.</p>

1. What is momentum and how is it related to pure rolling and trajectory physics?

Momentum is a measure of an object's motion, determined by its mass and velocity. In pure rolling, an object is rolling without slipping, meaning that its point of contact with the ground is stationary. In trajectory physics, momentum is important because it determines the path an object will take when it is launched into the air.

2. How is momentum conserved in pure rolling?

In pure rolling, momentum is conserved because there is no external force acting on the object. This means that the total momentum of the object before and after rolling remains the same. This is because the object's rotational and linear velocities are directly related to each other, and any changes in one will be offset by changes in the other to maintain the same total momentum.

3. What factors affect the trajectory of an object in motion?

The trajectory of an object is affected by its initial velocity, the angle at which it is launched, and the force of gravity. Air resistance can also play a role in altering the trajectory of an object.

4. How can we calculate the momentum of an object in pure rolling?

The momentum of an object in pure rolling can be calculated by multiplying its mass by its linear velocity. This is because the object's rotational and linear velocities are directly related, and the total momentum is the sum of both types of velocity.

5. Can momentum be transferred between objects in pure rolling?

Yes, momentum can be transferred between objects in pure rolling if there is an external force acting on the objects. For example, if two objects collide while rolling, the momentum of one object will be transferred to the other object. However, if there are no external forces, the total momentum of the system will remain constant.

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