Momentum -- pure rolling and trajectory physics problem

[timetraveller123]

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

 

[haruspex]

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

Consider the landing itself in terms of impulses. There is a vertical impulse mv_y. There will be a horizontal frictional impulse of J_x ≤ μ_smv_y.
The ball has a moment of inertia I about its centre. The frictional impulse generates a torque impulse J_xr. If this immediately leads to a rotation rate of ω then Iω=J_xr.
At the same time, the horizontal momentum changes from mv_x to mv_x-J_x.

Suppose that means an instant transition to rolling. What equation can you write?

 

[timetraveller123]

please give me some time my exams are going on i will repost once i have time thanks for your understanding

 

[timetraveller123]

yay finally got a break i am back !

ok i have been working in terms of impulses now using the defintions you used:mv_x - J_x = mv_i
w_i = J_x r/I

m(v_f - v_i) = mgμt
I(w_f - w_i) = mgμtr

v_f = (mv_x - J_x)/m + gμt
w_f = (mgμrt + J_xr)/I
then what do i do do ?

 

[haruspex]

yay finally got a break i am back !

ok i have been working in terms of impulses now using the defintions you used:mv_x - J_x = mv_i
w_i = J_x r/I

m(v_f - v_i) = mgμt
I(w_f - w_i) = mgμtr

v_f = (mv_x - J_x)/m + gμt
w_f = (mgμrt + J_xr)/I
then what do i do do ?

No, I said to consider the possibility that it transits immediately to rolling. So there is no t, v_f or w_f. if it is rolling straightaway, what is the relationship between v_i and w_i?

 

[timetraveller123]

oh then v_i = w_ir? why are you asking that

 

[haruspex]

oh then v_i = w_ir? why are you asking that

So combine that with your first two equations in post #34.

 

[timetraveller123]

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

 

[haruspex]

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.

 

[timetraveller123]

ok i kind of get what you are saying now so i did this:
no more vf and wf and t

mv_x = J_x = mw_ir
mw_ir = J_x r²m/I
solving for J_x i got
J_x = 2mv_x/7
then what the other equation we have is
j_x ≤ μJ_v am i supposed to use this
is what i did correct

 

[haruspex]

ok i kind of get what you are saying now so i did this:
no more vf and wf and t

mv_x = J_x = mw_ir
mw_ir = J_x r²m/I
solving for J_x i got
J_x = 2mv_x/7
then what the other equation we have is
j_x ≤ μJ_v am i supposed to use this
is what i did correct

Yes, that all looks good. It should give you the relationship between μ, v_x and v_y which allows instantaneous transition to rolling.

(I was thrown at first by a typo, = for -, in your first equation above.)

 

[timetraveller123]

oh you sorry about the typo
i don't quite get what you are saying after that we know v_x we know μ how are we even supposed to find time from that equations

 

[haruspex]

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?

 

[timetraveller123]

isn't that instantaneous?

 

[haruspex]

isn't that instantaneous?

Quite so, but you asked how we would find time.

 

[timetraveller123]

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!

 

[haruspex]

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 v_x, v_y and μ.

 

[timetraveller123]

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 J_x = J_vμ then there is instantaneous transition to pure rolling
so since J_x here is 2mv_x /7 ≤μmv_y hence there is immediate transition to pure rolling
is my understanding correct

 

[haruspex]

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 J_x = J_vμ then there is instantaneous transition to pure rolling
so since J_x here is 2mv_x /7 ≤μmv_y hence there is immediate transition to pure rolling
is my understanding correct

Yes, that's it.

 

[timetraveller123]

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

 

[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