1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Acceleration due to Impulse

  1. May 1, 2017 #1
    1. The problem statement, all variables and given/known data
    A uniform circular disc of mass 2m and radius R placed freely on a horizontal smooth surface as shown in the figure. A particle of mass m is connected to the circumference of the disc with a mass less string. Now an impulse J is applied on the particle in the directions shown in the figure. The acceleration of center of mass of the disc just after application of impulse is
    question.png

    2. Relevant equations

    $$F = ma$$



    3. The attempt at a solution
    assuming tension T in the string
    Force acting on particle = T
    Force acting on disc, ##T = 2ma## , a is the acceleration of center of mass of disc
    Torque on disc ##TR = Iq## , q is the angular acceleration
    hence ##TR = \frac {2mR^2} {2}q##
    thus ##T = mRq##
    also since impulse is change in momentum, velocity of particle, ##v= \frac{J}{m}##

    I cant seem to find a relation between velocity and tension. Does the particle follow a circular path? If so, do i take tension to be equal to the centripetal force?
     
  2. jcsd
  3. May 1, 2017 #2

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    What is the acceleration of the point on the disc where the string attaches?
     
  4. May 1, 2017 #3
    ##a + qR##
     
  5. May 1, 2017 #4

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Right, so how does that modify the equation of motion of the mass m?
     
  6. May 1, 2017 #5
    The problem looks strange. Initially no forces act on the disk in horizontal direction . Acceleration of the disk's center is equal to zero at initial moment of time.
     
  7. May 1, 2017 #6

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Not in the left-to-right direction, no. (It is all horizontal.)
     
  8. May 1, 2017 #7
    ##T - m(a + qR) = \frac{mv^2}{2R}##
     
  9. May 1, 2017 #8
    Pardon, I do not understand little bit.Do you agree that in the left-to-right direction there are no forces at initial moment?
     
  10. May 1, 2017 #9

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Yes, I already agreed with that. No acceleration, initially, in the left-to-right direction.
     
  11. May 1, 2017 #10
    Then it is a solution is not it?
     
  12. May 1, 2017 #11

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Not quite. That would imply the tension is even greater than if the disc were fixed.
     
  13. May 1, 2017 #12

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    No, what about the other horizontal direction, up and down the page?
     
  14. May 1, 2017 #13
    As I understand the statement of the problem the disk can not move up or down
     
  15. May 1, 2017 #14

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Not up or down in space, up or down in the page. It is a plan view.
     
  16. May 1, 2017 #15
    O now I see, the picture is a view from above on a table where the disk lies. Then it is a good problem
     
  17. May 1, 2017 #16
    But if i were to add the forces instead of subtracting them wouldn't it mean that the string would go slack?
     
  18. May 1, 2017 #17

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Only if the acceleration of the point of attachment exceeds v2/2R, which presumably it won't.
     
  19. May 1, 2017 #18
    I got confused...shouldn't ##ma + mqR## be because of the tension in the string and so, the net force acting on the particle is T which would provide centripetal acceleration.
     
  20. May 1, 2017 #19
    Actually this needs proof. It is a sole nontrivial point I guess.
     
  21. May 1, 2017 #20

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    But how much acceleration is needed? The standard formula v2/r assumes a fixed centre of rotation. In this case, it is accelerating towards the mass. The fixed quantity is the length of the string. The acceleration the mass needs to achieve that is reduced to the extent that the other end of the string is accelerating towards it.
     
  22. May 1, 2017 #21
    Finally got it! Thank you so much!
    ##a = \frac{J^2}{10m^2R}##
     
  23. May 2, 2017 #22
    Hi haruspex ,

    So , if the string length is constant , does this mean the particle is accelerating downwards (in the direction along the length of string away from disk ) with same acceleration as that of the point where string is attached to disk I.e ##a+qr## ?

    I tried approaching this problem mathematically using polar coordinates .

    Let the origin be a fixed point on the table just below the point where string leaves the disk and use polar coordinates ##r## and ##\theta## for the position of the particle. In polar coordinates the radial component of acceleration is ##a_r = \ddot{r} – r\dot{\theta}^2##

    If I use ## \ddot{r} = a+qR ## , the equation of motion of particle would be

    ## -T = m[(a+qR) - \frac{v^2}{2R}]## .

    This does give correct result :smile:

    Is my choice of origin correct ?

    Have I used correct value of ##\ddot{r}## considering the string constraint ? I am especially quite doubtful in this .Is particle actually accelerating downwards instantaneously with magnitude ( a+qR ) ?

    In case haruspex has signed off for the day , would @TSny look at my approach .

    Thanks
     
  24. May 2, 2017 #23

    TSny

    User Avatar
    Homework Helper
    Gold Member
    2017 Award

    Hi, Vibhor.
    Did you mean to write ##a+qR## here?

    The particle is not accelerating downward. The particle is accelerating upward (note the direction of the force on the particle).

    However, in your polar coordinate system, you can show that ##\ddot{r} = a+qR## just after the impulse. That is, ##\ddot{r}## does equal the magnitude of the acceleration of the point of attachment of the string to the disk.

    But ##\ddot{r}## does not represent the downward acceleration of the particle. The downward acceleration of the particle is ##\ddot{r} - r\dot{\theta}^2##.
     
  25. May 2, 2017 #24
    Could you please explain this .
     
  26. May 2, 2017 #25

    TSny

    User Avatar
    Homework Helper
    Gold Member
    2017 Award

    Let ##b## denote the point of attachment of the string to the disk and let ##p## refer to the particle. Let ##\hat{r}## be the radial unit vector in your polar coordinate system. ##\hat{r}## points downward at the time of interest.

    Relative acceleration formula: ##\vec{a}_p = \vec{a}_{p/b} + \vec{a}_b##

    At the time of interest
    ##\vec{a}_p = (\ddot{r} - r\dot{\theta}^2) \hat{r} \,\,\,## (1)
    ##\vec{a}_{p/b} = - r\dot{\theta}^2 \hat{r} \,\,\,\;\;\;\;## (2)

    To obtain(2), use the fact that the string can't stretch.

    Substitute (1) and (2) into the relative acceleration formula.
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted