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Solid sphere rolling down a house roof... angular speed

  1. Jul 3, 2017 #1
    1. The problem statement, all variables and given/known data
    A solid sphere of radius 16cm and mass 10kg starts from rest and rolls without slipping a distance of 9m down a house roof that is inclined at 43 degrees.

    What is the angular speed about its center as it leaves the house roof?

    The height of the outside wall of the house is 6m. What is the horizontal displacement of the sphere between the time in which it leaves the roof and the time at which it hits the ground?

    2. Relevant equations

    Torque=Inertia*Angular Acceleration (?)

    Inertia of a solid sphere =0.4*m*r^2

    I'm not really sure.. but:

    Vf^2=Vo^2+2ad

    3. The attempt at a solution

    Inertia of the sphere.. incase I need it?
    I=0.4*10kg*0.16m^2= 0.1024 kg*m^2

    Viy= 0m/s
    Vfy= ?
    Ay= -9.8m/s^2
    delta y= 9sin43= -6.1379m (vertical displacement)

    Then I tried to calculate for the final vertical velocity (before it rolls off the roof)
    Vfy^2=Viy^2+2ady

    Vfy= sqrt(2*-9.8m/s^2*-6.1379m)

    Vfy= 10.968 m/s

    Then I tried to calculate for the final velocity before the ball rolls off the roof:
    Vf= 10.968/sin43 = 16.0821

    Then I tried to convert it to angular velocity:
    V=r*ω
    ω=V/r
    ω=16.0821m/s / 0.16m = 100.5 rad/s


    I know this is incorrect, but I'm unsure as to where I went wrong. I'm trying to solve the problem without using work/energy, as my teacher hasn't yet covered those topics. If someone could point me in the right direction it would be GREATLY appreciated!!! Thanks :)
     
  2. jcsd
  3. Jul 3, 2017 #2

    BvU

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    Hi,

    Did you make a sketch of the various forces that accelerate/slow down the ball ?
    No. See your drawing. And use subscripts.
    Where do you use that ?
    Where do you use ##\tau = I\alpha## ?
    Who is ady ? Use only variables you have explained. Don't use A and a to denote the same variable.
     
  4. Jul 10, 2017 #3

    rude man

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    Basic hint: sum of kinetic energies due to distance traveled by the c.g. plus rotational kinetic energy = total loss of potential energy.
     
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