The problem involves two solid spheres of different masses colliding and subsequently moving in projectile motion and rolling. The original poster describes a scenario where one sphere comes to rest after the collision while the other continues its motion, eventually transitioning from projectile motion to pure rolling after hitting the ground.
The discussion is ongoing, with participants attempting to clarify their understanding of the physics involved. Some have offered guidance on the dimensional analysis of impulse and force, while others are exploring the conditions necessary for rolling motion to occur. There is no explicit consensus yet, as participants are still working through their interpretations and calculations.
Participants note potential confusion arising from the problem statement and the provided diagram, which may not accurately represent the scenario. There are also mentions of varying normal forces during the impact and the need for clarity on the height and velocities involved in the motion.
Consider the landing itself in terms of impulses. There is a vertical impulse mvy. There will be a horizontal frictional impulse of Jx ≤ μsmvy.vishnu 73 said:huh then what do i use for the friction once landing because i have become too used to writing f = μN and i haven't really done any problems whereby the full frictional force is not "used" would be helpful if you could guide me through this one thanks
No, I said to consider the possibility that it transits immediately to rolling. So there is no t, vf or wf. if it is rolling straightaway, what is the relationship between vi and wi?vishnu 73 said:yay finally got a break i am back !
ok i have been working in terms of impulses now using the defintions you used:mvx - Jx = mvi
wi = Jx r/I
m(vf - vi) = mgμt
I(wf - wi) = mgμtr
vf = (mvx - Jx)/m + gμt
wf = (mgμrt + Jxr)/I
then what do i do do ?
So combine that with your first two equations in post #34.vishnu 73 said:oh then vi = wir? why are you asking that
I'm not assuming it transits immediately to rolling. I am trying to find out if that is possible.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
Yes, that all looks good. It should give you the relationship between μ, vx and vy which allows instantaneous transition to rolling.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
If we can show the transition is instantaneous upon landing, how long does the transition take?vishnu 73 said:how are we even supposed to find time from that equations
Quite so, but you asked how we would find time.vishnu 73 said:isn't that instantaneous?
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.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!
Yes, that's it.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.vishnu 73 said:so before each problem i should check whether the impulse is enough to cause pure rolling instantly?
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.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?
No, as I wrote, it can be viewed either way. The total KE is the same.vishnu 73 said:so it is just that about the instantaneous center i only regard the rotational energy?
That's different. There is no rotational KE here since no rigid body is rotating.vishnu 73 said:then what is energy of the Earth around the sun ignoring Earth's spin on its own axis
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.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?
So, the ball is rolling to the right, say, and we yank the paper suddenly to the left.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?