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Disk rolling down a hill with no sliping

  1. Apr 22, 2012 #1
    A disk with mass m and radius R is rolling down a hill with no slipping, until it reaches a wall and then stops:


    zcmwywz2qy3n.png

    I want to write a set of equations describing the position cordinates x and y of the disk's center of mass (point A) and the cordinates for point B.

    mgsin(theta)=ma -> gsin(theta)=a -> gsin(theta)=dv/dt ; V_B=2wR ; V_A=wR

    Can you help me figure this out and guide me towards the solution?

    Thanks,
    Sharon
     
    Last edited: Apr 22, 2012
  2. jcsd
  3. Apr 22, 2012 #2

    ehild

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    The CM moves as if the whole mass was centred there, under the effect of all forces acting on the disk.
    You wrote the component of gravity along the slope as force, but there is an other one, which makes the disk rolling instead of slipping. What is it?

    You can also use conservation of energy to find out the velocity of the CM and angular velocity of the rotation of the disk.

    If you have figured out the motion of the CM, the point on the rim performs circular motion around the CM, with angular velocity vCM/R. You have to add the velocities ; those of translation and rotation.

    ehild
     
  4. Apr 22, 2012 #3
    You mean friction?
    Is it correct to say that my coefficient of friction is mew=1, giving me:
    mgsin(theta) - mgcos(theta) = ma ?

    using energy:
    mg(h+R)=0.5mv^2

    I still wonder how i am using these equations to create the connection between cordinates x,y of points A and B to time,mass,radius,height,angle ?
     
  5. Apr 22, 2012 #4
    Ignore my last message,

    1. -f+mgsinθ=mx'' , 2. N=mgcosθ Forces equations
    3. -fR=-0.5mR^2 * θ'' Momentum equation
    4. Rα''=-x'' Pure rolling condition

    After some algebra i get:

    5.1 x''=2/3 gsinθ
    5.2 α''=-2/3 * g/R * sinθ
    5.3 f=1/3 mgsinθ
    5.4 N=mgcosθ

    And as long as μ>1/3 tanθ there is a pure roll with no slip

    x'=2/3 gtsinθ+x'(0)=2/3 gtsinθ , x(t)=1/3 gt^2 sinθ+x(0)=1/3 gt^2 sinθ

    Now, how do I find x(t) and y(t) for points A and B ?
     
    Last edited: Apr 22, 2012
  6. Apr 22, 2012 #5

    ehild

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    It is static friction (Fs) between the disk and slope that does not let the disk slip. Static friction is not a force of defined magnitude: it has got the necessary magnitude. The static friction decelerates the translation of the CM, but accelerates rotation.

    You have two equations, one between forces and acceleration of CM:

    ma=mgsinθ-Fs

    the other for torque (FsR), moment of inertia (I) and angular acceleration dω/dt

    Idω/dt=R Fs,

    You have the condition of rolling Rω=VCM, which holds also for the accelerations Rdω/dt=a.
    Eliminate Fs and you get the acceleration a.

    ehild
     
  7. Apr 22, 2012 #6
    I posted another message, can you take a look at the equations I wrotre there?
     
  8. Apr 22, 2012 #7

    ehild

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    You have found x(t): it is the position of point A along the slope, till the disk is on the slope. It is not the x coordinate so it would be better to use s(t) instead of x(t). x(t)=s(t)cos(theta)
    You get the change of y as the y coordinate of the displacement, and add the initial height: yi=h+R

    For x>L1, the disks moves on the horizontal track with constant velocity.

    As for the point on the rim, write up its position in terms of the position of the CM and the position of B with respect to the CM.

    ehild
     

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  9. Apr 23, 2012 #8
    I got another question, this dosent add up but it seems like the mass plays no role in the motion. is that correct?
     
  10. Apr 23, 2012 #9

    ehild

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    It does not matter.

    ehild
     
  11. Apr 23, 2012 #10
    thanks you, you had been a great up!
    now im left with writing the x(t) equation for the center of mass while moving on the plane (x>L1), how do i approach this one?
    is it correct to say that now my x(t)=sqrt(2gH)*t + x(0)=sqrt(2g*(h+R)) + 0 ?
     
  12. Apr 23, 2012 #11

    ehild

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    The disk will roll further with the same speed and angular speed as it reached the bottom of the slope, and the change of height is h. The CM can not go deeper than the radius. The initial potential energy transforms to the sum of KE of the translation and KE of rotation.

    ehild
     
  13. Apr 23, 2012 #12
    the equations i wrote dont work for me could you tell me where i went wrong?
    what i did was
    mgh=0.5mv^2
    v=x'=sqrt(2gh)
    x=sqrt(2gh)t+x(0)=sqrt(2gh)t+L1
     
  14. Apr 23, 2012 #13

    ehild

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    That is not true.

    mgh is equal to the total kinetic energy: 1/2 mv2+1/2 Iω2.


    ehild
     
  15. Apr 23, 2012 #14
    Thanks,
    now i know that I=0.5mR^2, i know R,m and h
    though im not sure how do i find the ω ? is it v/R ?
     
  16. Apr 23, 2012 #15

    ehild

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    In case of pure rolling, it is.

    ehild
     
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