Momentum -- pure rolling and trajectory physics problem

[timetraveller123]

Homework Statement

Two solid spheres one of mass 'm' and other of mass '2m' are comin towards each other(translationally) with a velocity v1 and v2.After collision the sphere of mass 2m comes to rest and the sphere of mass m does oblique projectile motion at an angle 30 with the vertical till it reaches at a height "H"m wall.(no rotational till now). After reaching height H,it again goes for a horizontal projectile motion with initial velocity same as the final velocity of horizontal projectile motion.The time for horizontal projectile is 1/20sec.After reaching ground with a velocity v3(final velocity of horizontal projectile),it first acquires some w(omega) angular velocity and does pure rolling for some time. (Coefficient of friction is 1/squareroot(2) viz 1/1.414 on the left side if the wall).Find the time taken for pure rolling after reaching the ground on the left side of the wall?

Homework Equations

for pure rolling :wr = v
momentum : p_i = p_f
friction: f = N μ

The Attempt at a Solution

[/B]
ok i am stuck at right the first part it's either i am making a very dumb mistake or i am not seeing something
taking +x to be right
momentum initailly is 5m - 4m = 1m
then final momentum in x must also be + 1m
then since the 2m ball stops then the 1m ball must have a momentum of 1m ie it must be moving to the right and not to left as shown then how is it possible ? help thanks!

 

[haruspex]

momentum initailly is 5m - 4m = 1m

You are only told they were moving in opposite directions. You are not told which was going left and which right.

 

[timetraveller123]

it was given in the pic

 

[haruspex]

it was given in the pic

Didn't you draw that? Were you given a picture?
If you were not given that picture then I am suggesting your picture is wrong

 

[timetraveller123]

i was given but i have a feeling it is wrong i didn't draw that

 

[haruspex]

i was given but i have a feeling it is wrong i didn't draw that

Ok. The picture isbackwards. The 1kg ball should have been heading towards the wall.

 

[timetraveller123]

ok then i will give it a try again

 

[timetraveller123]

ok this is what i did but i am doing something wrong please highlight my mistake to me thanks.

the small ball is starting its trajectory on the right side with a horizontal velocity of 1 hence at the top of the wall when it only has horizontal velocity it has a velocity of 1 upon its descent by energy conservation it has vertical velocity before hitting the ground of 10 and horizontal velocity of still 1 . that is just before the collision.

hence after the collsion the ball starts rolling/sliding hence it does not bounce
thus,
Ndt = Δp = 10m N is normal force exerted by ground as the ball come to stop in vertical direction the change in momentum is 10m m is the mass of ball
let f be the frictional force from the ground which is Nμ
v_i be velocity of ball right after impact w_i be angular velocity of ball right after impact
v_f be velocity of ball after pure rolling has started and w_f be angular velocity of ball after pure rolling has started
let r be radius of ball and m be mass of ball
hence
fdt = Δp_x = m(v_i - 1) where 1 was the velocity of ball in horizontal direction before collision
and
frdt = ΔL= I(w_i - 0) as the ball was not spinning before collision and I is the moment of inertia of ball
substituting fdt = Ndtμ = 10 m μ
10μ + 1 = v_i and
10 m μ r = 2/5 m r₂w_i
w_i = 25μ/r
after some time t the ball will start pure rolling so we can again write but this time the N is m instead of 10m like before
ft = Δp_x = m(v_f - v_i) and
frt = ΔL = I(w_f - w_i ) and now we have
w_f = v_f/r
frt = 2/5 m r² ( v_f/r - 25μ/r)
mgμrt = 2/5 mr² ( v_f/r - 25μ/r)
gμt = 2/5(v_f - 25μ)
and using the other impuse equation we get
mgμt = m(v_f - v_i)
gμt = v_f - 10μ -1

we now have two equations for gμt namely
gμt = v_f - 10μ -1
and
gμt = 2/5(v_f - 25μ)
so solving for t using the μ given
i get t=-0.906 :(
what did i do wrong
thanks for your help!

 

[haruspex]

Ndt = Δp = 10m N is normal force exerted by ground

You are confusing impulse (change in momentum) with force. They are dimensionally different and cannot be equated. The normal force during landimg will be much higher. How high depends on how long it takes to halt the downward movement. On concrete the forces will be much higher than on rubber.

10μ + 1 = vi

This is even more confused dimensionally. It is very good style to work entirely symbolically, using just names for variables, not plugging in any numbers until the very end. It has many advantages, one being that it allows you to check whether your equations make sense dimensionally.

The problem statement is very strange. I have no idea what the platform on top of the wall is for. It seems it is smooth, so the ball gains no rotation and loses no linear speed while sliding along it.
Leaving that aside, the ball mass m lands from a height of h=10m, with a forward speed of v=1m/s, and no initial rotation. The impact generates an upward impulse of m√(gh).
Now, the thing about impacts between rough surfaces is that if there is any lateral motion of the bodies then there is a corresponding impulsive friction. If the static coefficient is μ, the maximum frictional impulse is μm√(gh). This could be enough to achieve rolling motion immediately. See if you can work out whether it is.

 

[timetraveller123]

i really am confused when it comes to rolling question i just tried to apply what i learned in this question
https://brilliant.org/problems/rolling-projectile-2/
see the solution posted to that it follows the same structure as mine
i don't understand what you mean by don't mix impulse and force aren't they related as impulse = f *t=change in momentum
and i don;t understand why is vertical impulse J_v = m√gh as vertical impulse = change in momentum in vertical which equals m(v_f - v_i) then J_h = μJ_v
and once again i really suck at rolling problems and please enlighten what is wrong with my understanding as not many book taught me pure rolling with collision explicitly thanks once again

 

[haruspex]

i really am confused when it comes to rolling question i just tried to apply what i learned in this question
https://brilliant.org/problems/rolling-projectile-2/
see the solution posted to that it follows the same structure as mine
i don't understand what you mean by don't mix impulse and force aren't they related as impulse = f *t=change in momentum
and i don;t understand why is vertical impulse J_v = m√gh as vertical impulse = change in momentum in vertical which equals m(v_f - v_i) then J_h = μJ_v
and once again i really suck at rolling problems and please enlighten what is wrong with my understanding as not many book taught me pure rolling with collision explicitly thanks once again

Sorry, I did make a few mistakes because I'd had trouble getting net connection and when I got it misremembered some of the data. I was thinking 10m was the height, e.g.
(It would help if you were to include units. Also got a bit confused by use of N as normal force, not N as Newtons. Besides, the normal force varies during impact, and we do not care what its value is. All we care about is the net change in momentum, so you do not need a variable to represent it.)
Re, m√(gh), that was a typo. I meant m√(2gh). But since we do not need to calculate h you can ignore that.
So, restart.

You seem to be saying it lands with a vertical speed of 10m/s. That is not right. It takes off with horizontal speed 1m/s and at 30^o to the vertical. That tells you its vertical launch speed. The landing speeds will be the same, if I am interpreting correctly.
Anyway, it is much better style not to plug in numbers until the end (and helps avoid confusing careless readers). So let's call the horizontal and vertical speeds on landing u and v respectively.

The vertical impulse is mv (no bounce), so the frictional impulse is anything up to μmv. The first question is whether that is enough to achieve rolling instantly. Don't worry about anything later just yet.
If the max frictional impulse occurs, the new horizontal speed will be ... What? You wrote 10μ+1, which I interpret as vμ+u. That is not right. Which way will the frictional impulse act?

 

[timetraveller123]

no because the question is vague i didn't whether to calculate final downward velocity as √2gh or the initial upward vertical velocity from the collision each of which yield a different final downward velocity hence which one should i use?

 

[haruspex]

no because the question is vague i didn't whether to calculate final downward velocity as √2gh or the initial upward vertical velocity from the collision each of which yield a different final downward velocity hence which one should i use?

Surely you had to calculate h from the upward velocity, and then use that to find the downward velocity on landing. So I do not understand how the two vertical speeds can be different. Please post your working.

 

[timetraveller123]

oh no that's not what i did i thought the question meant the vertical velocity from the collision was just enough to clear the wall that was why i got contradicting values so basically in this question the wall does not play any use because in √2gh i used the h of wall and not the highest point of trajectory as i thought the wall meant something so should i ignore the wall?

 

[haruspex]

oh no that's not what i did i thought the question meant the vertical velocity from the collision was just enough to clear the wall that was why i got contradicting values so basically in this question the wall does not play any use because in √2gh i used the h of wall and not the highest point of trajectory as i thought the wall meant something so should i ignore the wall?

The question certainly implies the top of the wall is the highest point of the trajectory, but isn't that how you found the height of the wall?
Several of your posts suggest to me that you think you know the height of the wall from some other information, but I do not see it given anywhere.

 

[timetraveller123]

ooh it is given in the picture the picture says H = 5m

 

[haruspex]

ooh it is given in the picture the picture says H = 5m

Well, I do not see it in the picture you posted, and anyway that conflicts with the other information. To reach that height it would have needed an upward speed of 10m/s after impact. But you have calculated its forward speed was 1m/s then, and we are told it rose at 30 degrees to the vertical. Not all of these things can be true.

 

[timetraveller123]

ok well then i will try again this time just using the vertical velocity from the collision and by the way it is in on the top right hand corner of the right page of the image where it says H = 5m on the left of g = 10m/s²

 

[timetraveller123]

sorry for the late reply had a lot of school work sorry :(
OK anyways back to question i keep getting negative values for t:

so the very initial horizontal velocity of ball after collision with the other bigger ball is 1(leaving out units for now)
so since the ball leaves at 30° to vertical the vertical velocity of ball is √3 in y direction
so the ball would be coming back down with √3 velocity downwards and 1 velocity to the left .
so again:
m be mass of ball
r be radius of ball
I be moment of inertial of ball = 2/5 m r²
f be friction acting on the ball
N be the normal force acting on the ball
v_i be initial translational velocity of ball after first impact and v_f be final velocity of ball after pure rolling has been achieved
w_i be initial rotational velocity of ball after first impact and w_f be final velocity of ball after pure rolling has been achieved
f = Nμ
Ndt = m√3
fdt = m(v_i - 1) ⇒v_i = √3 μ + 1
frdt = I (w_i) ⇒ w_i = 5√3 μ/(2r)

after the first collision the normal force acting on the ball reduces to just mg hence let t be time after which pure rolling is achieved
N = mg
f =mgμ
ft = m(v_f - v_i) ⇒ v_f = gμt + √3 μ + 1
and using w_f = v_f /r and subbing for what we got for v_f previously into this we can get the following
frt = I(w_f - w_i) ⇒ 5 gμt = 2μgt -3√3 μ + 2
3μgt = 2 - 3√3 μ
solving for t using μ = 1/√2 and g = 10
i get t = negative
don't know what i am doing wrong :(((
please help thanks!

 

[J Hann]

Apparently both spheres have to have the same radius since it is not given in the question.
Using that assumption, can't you get the final speed of m1 by just using conservation
of momentum (translational + rotational) when it reaches a pure rolling motion (stops slipping)?
If you can calculate that speed then it should be straightforward to
calculate the time required to reach the corresponding rate of rotation.

 

[haruspex]

Apparently both spheres have to have the same radius since it is not given in the question.
Using that assumption, can't you get the final speed of m1 by just using conservation
of momentum (translational + rotational) when it reaches a pure rolling motion (stops slipping)?
If you can calculate that speed then it should be straightforward to
calculate the time required to reach the corresponding rate of rotation.

The horizontal speed of the levitated sphere can be obtained by momentum conservation, regardless of radii, and vishnu did that in post#1. The trouble is that the question asks for the time taken, after landing, to transit to pure rolling - not the final velocity.
If the vertical impulse on landing is sufficiently great it will make that transition immediately.

Unforrtunately, the given information is contradictory. From the launch angle after the collision of the balls, the vertical speed on landing should be √3m/s, but then it says the height of the wall is 5m, making it 10m/s. Of course, it could be that even the lower number is enough that no time is taken.

 

[haruspex]

fdt = m(vi - 1)

Which way does friction act here? Will v_i be more or less than 1m/s to the left?

 

[J Hann]

The horizontal speed of the levitated sphere can be obtained by momentum conservation, regardless of radii, and vishnu did that in post#1. The trouble is that the question asks for the time taken, after landing, to transit to pure rolling - not the final velocity.
If the vertical impulse on landing is sufficiently great it will make that transition immediately.

Unforrtunately, the given information is contradictory. From the launch angle after the collision of the balls, the vertical speed on landing should be √3m/s, but then it says the height of the wall is 5m, making it 10m/s. Of course, it could be that even the lower number is enough that no time is taken.

I was not thinking of the speed of the "levitated" sphere.
The initial angular momentum (about the point where the ball attains pure rolling motion) is entirely due
to the translational momentum of that ball.
The final angular momentum (when the ball stops sliding) is the due to the rotational + the translational momentun
of that ball about the point of contact of that ball. Momentum should be conserved and should provide the
the speed of the ball when it stops sliding. The problem is highly hypothetical - for the ball to immediately
stop sliding on contact would require more torque than the coefficient of friction could supply. Also,
to assume no bouncing is another problem. It appears that problem requires a number of assumptions
that are not given.

 

[haruspex]

I was not thinking of the speed of the "levitated" sphere.

Why not? That has the same as the horizontal speed with which it takes off and lands, which is all important.

for the ball to immediately stop sliding on contact would require more torque than the coefficient of friction could supply

No, that's a common mistake.
If the vertical impulse on landing is mv then there is a corresponding frictional impulse of (up to) μ_smv. That could be enough to transit to rolling immediately. Follow the link in post #10 for a worked example.

 

[timetraveller123]

Which way does friction act here? Will v_i be more or less than 1m/s to the left?

wait v_i is lesser than 1 then should i be doing ft = m(1-v_i)

 

[haruspex]

wait v_i is lesser than 1 then should i be doing ft = m(1-v_i)

Right.

 

[timetraveller123]

it still doesn't work because the only equation we need to change are the equations with just the translational velocity so here how i did it again,
ndt = m√3
fdt = m(1-v_i) ⇒ v_i = 1- √3μ (already negative but going on anyways)
w_i (remains the same as before ) = 5√3/(2r)

now
n = mg
then frt = (v_f /r - w_i) ⇒ 5gμt = 2v_f - 5√3 μ
ft = m (v_i - v_f) (as v_f is once again lesser ) ⇒ v_f = 1 - √3μ - mgμt
solving the two equation once again i get a negative t = -0.133s

 

[haruspex]

it still doesn't work because the only equation we need to change are the equations with just the translational velocity so here how i did it again,
ndt = m√3
fdt = m(1-v_i) ⇒ v_i = 1- √3μ (already negative but going on anyways)
w_i (remains the same as before ) = 5√3/(2r)

now
n = mg
then frt = (v_f /r - w_i) ⇒ 5gμt = 2v_f - 5√3 μ
ft = m (v_i - v_f) (as v_f is once again lesser ) ⇒ v_f = 1 - √3μ - mgμt
solving the two equation once again i get a negative t = -0.133s

Remember that the coefficient of static friction only provides an upper limit on the frictional force.

 

[timetraveller123]

what do you mean i don't understand what it means to this problem and doesn't this question require the use of kinetic friction until the ball starts pure rolling

 

[haruspex]

what do you mean i don't understand what it means to this problem and doesn't this question require the use of kinetic friction until the ball starts pure rolling

Kinetic friction will not occur unless the max static friction is exceeded at some point. You have applied max static friction to the horizontal impulse on landing and got the impossible result that the ball immediately starts moving to the right. What do you conclude?

 

[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?