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Help Challenging rolling ball problem ahead!

  1. Jan 31, 2005 #1
    Help!! Challenging rolling ball problem ahead!

    A solid sphere of rotational inertia I=2/5(MR^2) is rollling without slipping on a horizontal rough surface toward a rough inclined plane. The sphere's mass is 2kg and radius is 8 cm. The velocity of the sphere as it aproaches the incline is 10m/s.

    Since I don't know how to put the incline here, well, it has an angle of elevation of 30, and the distance of the hypothenuse of such triangle is 12.

    a) With what speed will the sphere reach the top of the incline assuming nos slippage on the incline either?

    I'd appreciate if someone explains in detail this problem. Thx.
    Last edited: Jan 31, 2005
  2. jcsd
  3. Jan 31, 2005 #2

    use conservation of energy....
    what is the KE b4 the ball rolling up.... KE=KE(rotational)+KE(translational)
    what is the PE at the top of hill?
  4. Jan 31, 2005 #3
    I dont know if this would work but try finding the kinetic energy of the shpere .5mv^2 then subtract the energy lost in potential mg(sin30)12 then that should equal the kinetic energy at the top of the slope then solve for the new v. Thats if the angular velocity stays the same.
  5. Feb 1, 2005 #4


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    The kinetic energy of the sphere consists of two parts:
    Translational kinetic energy associated with motion of the center of mass. ([itex]1/2mv^2[/itex]).
    Rotational kinetic energy associated with motion about the center of mass. ([itex]1/2I\omega^2[/itex])

    Since the ball rolls without slipping, [itex]v[/itex] and [itex]\omega[/itex] are related through the radius of the sphere.
  6. Feb 1, 2005 #5


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    But angular velocity CAN'T stay the same. The ball is rolling without slipping so the angular velocity has to reduce to match the linear velocity.
  7. Feb 2, 2005 #6
    As the sphere goes up the incline, it is exerted upon by three forces: its weight, the reaction force from the incline and the friction force responsible for rolling.

    As it goes up, it gains potential energy. The friction force does no net work because the point of application does not move (pure rolling means that [itex]v_{CM}[/itex] and [itex]\omega[/itex] are related through the radius of the sphere as mentioned earlier). The total energy however is constant. So you can use energy balance at the top (and for that matter any height) and the bottom of the incline (the parameters at the bottom are known so you have effectively one unknown, [itex]v_{CM}[/itex]).

    If you have a problem doing an energy balance, then I suggest you take another look at the problem as follows. (I will leave the math and computations to you...please do it...).

    Total Kinetic Energy of body at any time = [itex]\frac{1}{2}I_{CM}\omega^2 + \frac{1}{2}Mv_{CM}^2[/itex]

    Translational Component = [itex]\frac{1}{2}Mv_{CM}^2[/itex]
    Rotational Component = [itex]\frac{1}{2}I_{CM}\omega^2[/itex]

    You can apply the work-energy theorem separately for these components:

    Work done by net force acting on the body
    = [itex](mg\sin\theta-f)s_{CM} = \frac{1}{2}Mv_{CM}^2[/itex]

    Work done by net torque acting about the center of mass
    = [itex]fs_{CM} = \frac{1}{2}I_{CM}\omega^2[/itex]

    Here [itex]s_{CM}[/itex] is the distance traveled by the center of mass up the incline, [itex]\theta[/itex] is the angle of the incline. Relate these two and add the equations together. The resulting equation is the energy balance restated.

    To complete the solution, you must relate [itex]v_{CM}[/itex], [itex]\omega[/itex] and [itex]\alpha[/itex] with the radius of the sphere. (This is a constraint valid only for rolling.)

    Finally solve for [itex]v_{CM}[/itex].

    Hope this helps...

    Last edited: Feb 2, 2005
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