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Homework Help: Impulse Problem -- A rolling cylinder collides with a cylinder at rest

  1. Mar 19, 2015 #1
    1. The problem statement, all variables and given/known data
    A solid cylinder of radius R is rolling without slipping on a rough horizontal surface. It collides with another identical cylinder which is initially at rest on the surface. The coefficient of restitution for the collision is 1.The coefficient of friction between the cylinders and between each cylinder and the ground is ##\mu##.If the final angular velocity of the second cylinder after it starts pure rolling is ##a/b## where a and b are co prime positive integers find a+b.

    Given: Initial velocity of center of mass of the first cylinder is 17.5 m/s, R=1m, ##\mu##=0.5

    2. Relevant equations
    ##J=m(vf-vi)##
    ##L_{J}=I(\omega_{f}-\omega_{i})##

    3. The attempt at a solution

    I think the main challenge is to find the velocity of the CoM and the angular velocity of the second cylinder after collision. Then the final angular velocity can be found by conserving angular momentum about the point of contact of cylinder and ground.

    I began in his way-----

    I made the Free Body diagrams
    impulse.png

    Let the initial velocity of the CoM (of cylinder 1) of the cylinder and initial angular velocity of the cylinder be ##v## and ##\omega## respectively. Let ##\omega## be in clock wise direction initially.

    Let the final velocity of the CoM of (of cylinder 1) the cylinder and final angular velocity of the cylinder be ##v_{0}## and ##\omega_{0}## respectively. Let ##\omega_{0}## be in clock wise direction.

    Let the final velocity of the CoM (of cylinder 2) be ##v_{1}## and angular velocity be ##\omega_{1}## in anti-clock wise direction.

    As net impulse is equal to change in momentum and net angular is equal to the change in angular momentum so
    ## J=m(v_{0}+v)##.....................(1)
    ##\mu JR=I(\omega_{0}-\omega)##.........................(2)

    For cylinder 2
    As the cylinder don't jump so
    ##\mu J=J_{N}## (I have neglected mg as it is very small as compared to the impulses)

    As net impulse is equal to change in momentum and net angular is equal to the change in angular momentum so
    ##J-\mu J_{N}=mv_{1}##
    ##I\omega_{1}=(\mu J-\mu J_{N})R##

    If we consider the time period of the collision is very small so I think we can neglect work done by friction during the collision and as the collision is elastic so I think we can conserve energy. So

    ##\frac { 1 }{ 2 } \left( m{ v }^{ 2 }+I{ \omega }^{ 2 } \right) =\frac { 1 }{ 2 } \left( m{ v }_{ 0 }^{ 2 }+m{ v }_{ 1 }^{ 2 }+I{ \omega }_{ 0 }^{ 2 }+I{ \omega }^{ 2 }_{ 1 } \right) ##

    Are my all equations correct?
     
  2. jcsd
  3. Mar 19, 2015 #2

    haruspex

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    Are you measuring v and v0 in opposite directions?
    For energy, it's not the time that matters but the integral of the force over the distance. ##\Delta E = \int F.ds = \int F.v.dt##, where v is the relative speed of the surfaces in contact. Taking v to be constant we get ##\Delta E = v\int F.dt = v\Delta J##. So the question is, is the relative velocity zero?
     
  4. Mar 19, 2015 #3

    TSny

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    Coefficient of restitution is often defined as the ratio of relative velocity of separation after the collision to relative velocity of approach before the collision. The velocities of separation and approach are components of velocity along the line of impact. If this definition is applicable here, then you don't need to worry about energy equations.

    See for example http://www.brown.edu/Departments/Engineering/Courses/En4/notes_old/oblique/oblique.html
     
  5. Mar 20, 2015 #4
    In case of extended bodies we have to consider the velocities of the point of contact of bodies. Right?
    As the collision is elastic so we can equate the velocity of approach and separation in common normal direction.
    So ##v=v_{0}+v_{1}##

    I have a confusion. In this situation impulses are acting not only due to interaction between the particles but also from external sources like impulse from the ground and friction. Can we use this concept even in these cases?:confused:
     
    Last edited: Mar 20, 2015
  6. Mar 20, 2015 #5
    Yes they are in opposite direction. Sorry, I didn't mention it.:sorry:

    In this (https://www.physicsforums.com/threads/just-a-confusion-a-mass-falling-onto-a-pivoting-rod.790098/) thread we solved the question by conserving angular momentum about the point O even if net external torque was acting on it. We equated the angular momentum of the system just before and after the collision. And we assumed that net torque didn't bring much change in the angular momentum of the system during collision. Why can't we do same thing with energies?

    Why can't I equate the total energy of the system before and after collision and assume the work done by the friction to be very small?:confused:
     
  7. Mar 20, 2015 #6

    haruspex

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    In that case you considered a limited, finite torque (MgL) acting for an infinitesimal time. The integral is zero. The torque of the impact (##mL\frac{\Delta v}{\Delta t}##)was unlimited, in the sense that the shorter the time it was considered to act over the larger the torque. In that case the integral is known and nonzero.
    In the present case, looking at the integral wrt distance, you have the same issue - a force which is unlimited. So it is not clear whether its integral over an infinitesimal distance will be zero.
    In attempting to solve the question myself, I used the same principle that TSny suggests. However, oblique impacts of elastic bodies is a complicated matter. Ever played with a superball? A spun ball dropped on concrete will not only bounce up at an angle, its direction of spin reverses. So not only does the linear motion bounce, the rotational motion does too. But in that case there is no slipping. It is not made clear, but it seems to me we are supposed to assume static friction is overcome.
     
  8. Mar 20, 2015 #7
    Thank you I got it now.:smile:
     
  9. Mar 20, 2015 #8
    I am getting ##{ v }_{ 1 }=\frac { 6\times 17.5 }{ 7 } ##
    and ##{ \omega }_{ 1 }=\frac { 4\times 17.5 }{ 7 } ##

    and final angular velocity when it starts pure rolling as
    ##{ \omega }_{ final }=\frac { 20 }{ 3 } ##

    Am I right?
     
  10. Mar 20, 2015 #9

    TSny

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    That's what I get using the relative velocity interpretation of the coefficient of restitution.
     
    Last edited: Mar 20, 2015
  11. Mar 20, 2015 #10
    Just a last question:
    Can we use the concept of coefficient of restitution in any case if value of ##e## is provided? I am asking this because I used to think that we can not use this when impulses acts from external sources like impulse due to friction and normal in this case.
     
  12. Mar 20, 2015 #11

    TSny

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    I have never seen a textbook problem where you cannot use ##e## in terms of relative velocities, even when there are external impulses during the collision. But I don't have the knowledge to make a general claim.
     
  13. Mar 20, 2015 #12

    haruspex

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    Usually the friction is orthogonal to the relative velocity, so it's ok. But here I think there's a problem. The friction between the second ball and the ground is also horizontal.
    It would be interesting to see what happens with an energy approach. I believe I can show that if an impulsive frictional torque (i.e. sudden change in angular momentum) JFr is applied to a body with MoI of I then the total work done is ##\frac{(J_Fr)^2}{I}##, but only half ends up as rotational KE.
     
  14. Mar 20, 2015 #13
    Thank you! :smile:
     
  15. Mar 20, 2015 #14
    Could you please tell me how you arrived at that result? Is ##J_F## is frictional force between cylinders?
     
  16. Mar 21, 2015 #15

    haruspex

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    I was using JF as a generic frictional impulse.
    But on second thoughts, I doubt my result is right, since it ignores the concurrent force from the impact of the first ball.
    I'll try again, but I'm a bit busy at the moment.
     
  17. Mar 21, 2015 #16
    No problem.:smile:
    My solution is wrong. In this I have assumed that limiting value of friction acts between cylinder and cylinder 1 slips over cylinder 2. However we can't do so. One of my friend has shown that friction between cylinders is sufficient for cylinder 1 to roll over cylinder 2. He got final ##\omega=455/48##. I am convinced with his solution. What do you guys think?
     
  18. Mar 21, 2015 #17

    TSny

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    It seems to me that the coefficient of friction given in the problem must be for kinetic friction. You need the coefficient of kinetic friction to determine how far cylinder 2 travels before it starts rolling without slipping. A coefficient of restitution greater than zero implies that the cylinders separate after the collision. So, I don't see how cylinder 1 is going to roll up over cylinder 2. Maybe I'm overlooking something.
     
  19. Mar 21, 2015 #18

    haruspex

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    As TSny has posted, your friend's solution is almost surely wrong.
    I've come to the conclusion that as far as the elastic collision is concerned the effect of the impulsive friction from the ground on the second cylinder is to make it seem more massive. Since the ratio of the relative speeds equation does not depend on the masses, it is still valid. So I agree with your answer in post #8.
     
  20. Mar 22, 2015 #19
    As there is no information so I think we should use ##\mu_{k}=\mu_{s}=0.5##.
    The e is greater than 0. So this means that they do stick with each other. But it may be possible that time period in which they(cylinders) remain in contact, they might roll over each other instead of sliding on one other. I think they can roll over each other for the small time in which they remain in contact with each other.
    What do you think guys?
     
  21. Mar 22, 2015 #20

    haruspex

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    It is quite impossible for the first cylinder to roll over the second. The instant after impact, it cannot have a greater horizontal speed than the second since that would mean the two intersect. Since it will become airborne, the two therefore separate.
    It is true that the second cylinder will now have backspin, so will lose some of its horizontal speed, so perhaps the first can catch it up, but I very much doubt it.
     
  22. Mar 23, 2015 #21
    I agree that cylinder 1 will not roll on 2 forever. I am saying that cylinder 1 can roll over cylinder 2 during small time period in which collision occurs if the friction is sufficient.
    I think I can explain this through a question--
    "A kid kicks a ball to the wall so that it rolls without slipping. After the collision with the wall, the ball will jump/kick up and make an angle with the ground. Estimate that angle in degrees, if the coefficient of friction between the ball and the wall is very high."
    Here in this question the ball when collides with the wall then impulse (due to friction)acts on the ball in upward direction. This impulse is responsible for vertical component of the velocity of the ball after collision. The angular impulse due to friction is also responsible for decreasing the angular velocity. Now as it is given that the friction between the ball and the ground is
    sufficient so ball tends to roll over the wall during the time of collision instead of slipping on it. The ball will definitely loose contact with the wall due to the horizontal impulse acting on it from the wall but the ball will roll over the wall during the time in which collision occurs. Similarly cylinder 1 will roll over cylinder 2 at the time of collision. But as collision ends then the cylinders will loose contact with each other and will definitely not roll over each other.
     
  23. Mar 23, 2015 #22

    haruspex

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    Maybe I misunderstand what you mean by its rolling over the other cylinder. As I explained, the impacting cylinder will come out of the collision with a lower horizontal velocity than the impacted cylinder, so although it will jump up, no part of it, initially at least, will be directly above any part of the the second cylinder. (As I wrote, what happens later as the backspin on the second cylinder takes effect I'm not sure.)
     
  24. Mar 23, 2015 #23
    I meant that during collision the velocities of their contact points are equal(i.e cylinder 1 would roll over cylinder 2). After collision they will not remain in contact and hence that condition will not be satisfied.

    I have raised this topic early too--
    Please have a look at the #post 47 and TSny's #post 48 https://www.physicsforums.com/threads/collision-of-a-particle-with-a-thin-rod.792356/page-3.
     
  25. Mar 23, 2015 #24

    haruspex

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    Oh, you mean rolls against, i.e. instantaneous rolling contact.
    Excellent point. Yes, I believe you are right. I get that if mu is the coefficient of friction between cylinder and ground then it will be rolling contact provided the coefficient between the cylinders is at least 2/(10-3 mu), which it will be. Can you confirm that calculation?
     
  26. Mar 24, 2015 #25
    Yes I was talking about instantaneous rolling contact.:woot:

    My friend got ##J=\frac{17mv_{0}}{16}## and ##J_{f}=\frac{mv_{0}}{4}##
    From here limiting value of ##J_{f}=\frac{17mv_{0}}{32}##
    As ##\frac{17mv_{0}}{32}> \frac{mv_{0}}{4}##. So the ball rolls.

    Are you feeling excited about 26 March(Australia vs India Semi-finals)?:nb):smile:
     
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