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Momentum -- pure rolling and trajectory physics problem

  1. Apr 14, 2017 #1
    1. The problem statement, all variables and given/known data
    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???
    0b0d4ea25f5a547dee3115cba62130db433f921c.jpg

    2. Relevant equations
    for pure rolling :wr = v
    momentum : pi = pf
    friction: f = N μ


    3. The attempt at a solution
    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!
     
  2. jcsd
  3. Apr 14, 2017 #2

    haruspex

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    You are only told they were moving in opposite directions. You are not told which was going left and which right.
     
  4. Apr 14, 2017 #3
    it was given in the pic
     
  5. Apr 15, 2017 #4

    haruspex

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    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
     
  6. Apr 15, 2017 #5
    i was given but i have a feeling it is wrong i didn't draw that
     
  7. Apr 15, 2017 #6

    haruspex

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    Ok. The picture isbackwards. The 1kg ball should have been heading towards the wall.
     
  8. Apr 15, 2017 #7
    ok then i will give it a try again
     
  9. Apr 16, 2017 #8
    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
    vi be velocity of ball right after impact wi be angular velocity of ball right after impact
    vf be velocity of ball after pure rolling has started and wf be angular velocity of ball after pure rolling has started
    let r be radius of ball and m be mass of ball
    hence
    fdt = Δpx = m(vi - 1) where 1 was the velocity of ball in horizontal direction before collision
    and
    frdt = ΔL= I(wi - 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 = vi and
    10 m μ r = 2/5 m r2wi
    wi = 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 = Δpx = m(vf - vi) and
    frt = ΔL = I(wf - wi ) and now we have
    wf = vf/r
    frt = 2/5 m r2 ( vf/r - 25μ/r)
    mgμrt = 2/5 mr2 ( vf/r - 25μ/r)
    gμt = 2/5(vf - 25μ)
    and using the other impuse equation we get
    mgμt = m(vf - vi)
    gμt = vf - 10μ -1

    we now have two equations for gμt namely
    gμt = vf - 10μ -1
    and
    gμt = 2/5(vf - 25μ)
    so solving for t using the μ given
    i get t=-0.906 :(
    what did i do wrong
    thanks for your help!
     
  10. Apr 16, 2017 #9

    haruspex

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    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.
    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.
     
  11. Apr 17, 2017 #10
    i really am confused when it comes to rolling question i just tried to apply what i learnt 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 Jv = m√gh as vertical impulse = change in momentum in vertical which equals m(vf - vi) then Jh = μJv
    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
     
  12. Apr 17, 2017 #11

    haruspex

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    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 30o 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?
     
  13. Apr 18, 2017 #12
    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?
     
  14. Apr 18, 2017 #13

    haruspex

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    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.
     
  15. Apr 18, 2017 #14
    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?
     
  16. Apr 18, 2017 #15

    haruspex

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    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.
     
  17. Apr 19, 2017 #16
    ooh it is given in the picture the picture says H = 5m
     
  18. Apr 19, 2017 #17

    haruspex

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    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.
     
  19. Apr 19, 2017 #18
    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
     
  20. Apr 20, 2017 #19
    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
    vi be initial translational velocity of ball after first impact and vf be final velocity of ball after pure rolling has been achieved
    wi be initial rotational velocity of ball after first impact and wf be final velocity of ball after pure rolling has been achieved
    f = Nμ
    Ndt = m√3
    fdt = m(vi - 1) ⇒vi = √3 μ + 1
    frdt = I (wi) ⇒ wi = 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(vf - vi) ⇒ vf = gμt + √3 μ + 1
    and using wf = vf /r and subbing for what we got for vf previously into this we can get the following
    frt = I(wf - wi) ⇒ 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!
     
  21. Apr 20, 2017 #20
    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.
     
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