Momentum -- pure rolling and trajectory physics problem

[haruspex]

so before each problem i should check whether the impulse is enough to cause pure rolling instantly?

Yes.

 

[timetraveller123]

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?

 

[haruspex]

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.

 

[timetraveller123]

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?

 

[haruspex]

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.

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.

 

[timetraveller123]

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?

 

[haruspex]

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 ∑½m_iv_i², 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 ∑½m_iv_i²=½(∑m_i)v_com²+½I_xω², where ω is the rate of rotation and I_x is a constant describing the distribution of mass within the body about the given axis. Specifically, I_x=∑m_ix_i², where x_i 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 ∑½m_iv_com²=½Mv_com².

 

[timetraveller123]

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

 

[haruspex]

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?

 

[timetraveller123]

ok say we pull it slowly then what happens

 

[haruspex]

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.

 

[timetraveller123]

ok thanks helped a lot that was all the doubts i had thanks once again

 

[neilparker62]

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.

 

[haruspex]

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.