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Sphere rolling down an incline problem.

  1. Nov 5, 2014 #1
    1. The problem statement, all variables and given/known data
    A uniform solid sphere, of radius 0.20 m, rolls without slipping 6.0 m down a ramp that is inclined at 28° with the horizontal. What is the angular speed of the sphere at the bottom of the slope if it starts from rest?

    2. Relevant equations
    KE=1/2Iw^2
    PE=mgh
    I don't know what to use for this.

    3. The attempt at a solution
    Ma = MgSin(28)-Ff
     
  2. jcsd
  3. Nov 5, 2014 #2

    BvU

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    Hi veg, welcome to PF.
    Your attempt at a solution has fallen off the map, and there is some more needed under 2) relevant equations.

    I hope you've done a few slip/non slip slope exercises and are familiar with the combination of rolling and accelerating ?

    In that case you know that he kinetic energy equation you listed applies to rotation about a stationary axis. Here, however, the axis is moving down the slope, which accounts for a kinetic energy due to linear motion too: KE = 1/2 m v2. So revise the list of eqns and show your attempt at solution , please :)
     
  4. Nov 5, 2014 #3
    1. The problem statement, all variables and given/known data
    A uniform solid sphere, of radius 0.20 m, rolls without slipping 6.0 m down a ramp that is inclined at 28° with the horizontal. What is the angular speed of the sphere at the bottom of the slope if it starts from rest?

    2. Relevant equations
    w=(theta)/t
    a=w/t
    v=r(theta)/t
    I don't know what to use for this.

    3. The attempt at a solution
    Acceleration = m(9.81m/s^2)(sin(28)) - Friction
    At the end it did 6.0m/2pi(0.20m) revs over the time it took. (4.78rev/t)
    from mgh=1/2mv^2+1/2Iw^2 i can get v^2=2gh/(1+I/mr^2)
    I have no idea how to find the time from these variables.
     
    Last edited: Nov 5, 2014
  5. Nov 5, 2014 #4

    BvU

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    Much better!
    This way I can ask some pertinent questions that might help you to get going.

    ##\omega =\Delta \theta/t## is for average angular velocity. Not good here: we have some constant acceleration and some constant angular acceleration, and the coupling is established by the non-slipping condition, which you kind of mention: ##v = \omega r##.

    Look up some equations for linear motion with constant acceleration plus some (surprisingly similar-looking) equations for angular motion with constant angular acceleration and see what you can use to get a grip on this nice exercise !

    There might even be a shortcut possible: the time isn't asked for, so perhaps an energy balance (more equations still!) gets us to the answer in one fell swoop (provided it's a complete balance...) !
     
  6. Nov 5, 2014 #5
    Ok i think i got it.
    v=rw
    x=r(theta)
    a=r(angular a)

    (9.81)(sin(28))/(0.20)=angular a=23
    (theta)=6/0.20=30
    w^2=2(23)(30)
    [w=37.1] is this correct?
     
  7. Nov 5, 2014 #6

    BvU

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    Acceleration = m(9.81m/s^2)(sin(28)) - Friction was better ! So ##\alpha## is alittle smaller than you think.

    Friction is the one that causes the ball to rotate around its axis. Look up ##\tau = I\alpha## and ##\tau = r \times F##.

    At the risk of repeating myself:
    Look up some equations for linear motion with constant acceleration plus some (surprisingly similar-looking) equations for angular motion with constant angular acceleration and see what you can use to get a grip on this nice exercise !
     
  8. Nov 5, 2014 #7
    Ok so what i did to try to find it is i took sin(28)*6 to find the height of at the start which is 2.82m
    Then i did mgh=1/2mv^2 + 1/2Iw^2 which worked out to be 2gh=v^2
    I got v as 7.438m/s and from that i did v/r=w
    so i got w as 37.2m/s
    I just need to convert it to rad/s
     
    Last edited: Nov 5, 2014
  9. Nov 6, 2014 #8

    BvU

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    ##mgh = {1\over 2} mv^2 + {1\over 2} I\omega^2\ \Rightarrow \ 2gh = v^2 + I\omega^2/m## is good. That's the energy balance.


    The ##2gh=v^2## that you worked out misses the rotation part and can't be right.
    And then: re dimension: ##v=\omega r##, so with v in m/s and r in m, the quotient is in radians/s, not m/s.


    Since you have ##v=\omega r## you only need ##I## to crack this one. (No need to go via ##v## if they want ##\omega##)
     
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