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Acceleration, Uniform Ball on Incline

  1. Dec 13, 2014 #1
    1. The problem statement, all variables and given/known data
    424c8aec49.8aebe7f7db.aACpWO.png

    A uniform solid ball of mass [itex]m[/itex] rolls without slipping down a right angled wedge of mass [itex]M[/itex] and angle [itex]θ[/itex] from the horizontal, which itself can slide without friction on a horizontal floor. Find the acceleration of the ball relative to the wedge.
    2. The attempt at a solution
    Attempt using generalized coordinates:
    Let us denote the coordinate of the ball as [itex]ξ[/itex] (which is along the incline) and the coordinate of the wedge as [itex]η[/itex]. The kinetic energy of the system is:

    [tex] K=\frac{1}{2}m\dotξ^2+\frac{1}{2}Iω^2+\frac{1}{2}M\dotη^2=\frac{1}{2}m\dotξ^2+\frac{2}{10}m\dotξ^2+\frac{1}{2}M\dotη^2=\frac{7}{10}m\dotξ^2+\frac{1}{2}M\dotη^2 [/tex]
    And the change in potential energy is:
    [tex] V=-mgξ\sin{θ} [/tex]
    The sum of both energies is the total energy, which is constant in time. We take the derivative of it:
    [tex] \dot{E}=\frac{d}{dt}E=\frac{d}{dt} K+V=\frac{d}{dt} (\frac{7}{10}m\dotξ^2+\frac{1}{2}M\dotη^2-mgξ\sin{θ})=\frac{7}{5}m\dotξ\ddotξ+M\dotη\ddotη-mg\dotξ\sin{θ}=0 [/tex]
    Finally, we need to relate both coordinates. Since the center of mass of the system doesn't move we have:
    [tex]mξ\cos(θ)=Mη[/tex] [tex]\frac{mξ\cos(θ)}{M}=η[/tex]

    We substitute [itex]η[/itex] in our derivative and we get:
    [tex]\dot{E}=\frac{7}{5}m\dotξ\ddotξ+\frac{m^2\dot{ξ}\ddot{ξ}\cos^2{θ}}{M}-mg\dot{ξ}\sin{θ}=0 [/tex]
    We solve for [itex]\ddotξ[/itex]:
    [tex]\ddotξ =\frac{5Mg\sin{θ}}{7M+5m\cos^2{θ}}[/tex]
    Can someone please confirm if this solution is correct?
     
    Last edited: Dec 13, 2014
  2. jcsd
  3. Dec 14, 2014 #2

    haruspex

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    Not exactly sure how you're defining ξ. The potential energy expression implies it's along the slope, relative to the wedge, but the KE expression implies it is in the instantaneous direction of linear movement of the ball relative to the ground.
     
  4. Dec 14, 2014 #3
    It is along the slope, How did the kinetic energy expression make you assume it's relative to the ground?
     
  5. Dec 14, 2014 #4

    haruspex

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    ##\frac{1}{2}m\dotξ^2##
     
  6. Dec 14, 2014 #5
    So the expression should have been: [itex] \frac{1}{2}m({{\vec{\dotξ}}}+{{\vec{\dotη}}})^2 [/itex]?
     
    Last edited: Dec 14, 2014
  7. Dec 14, 2014 #6
    Is the kinetic energy of the ball the sum of the kinetic energies of both coordinates? (ball moving with respect to the incline and the ground)

    [tex] \frac{1}{2}m\dotξ^2 +\frac{1}{2}m\dotη^2 [/tex]
     
    Last edited: Dec 14, 2014
  8. Dec 14, 2014 #7

    ehild

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    No.The translational KE of the ball is 1/2 m(vx2+vy2), vx and vy being the horizontal and vertical velocity components with respect to the ground. Express them with the velocity of the wedge and the velocity of the ball with respect to the wedge. Note that ξ has both horizontal and vertical components.
     
  9. Dec 14, 2014 #8
    The [itex]x[/itex] component of the ball's velocity with respect to the ground is:
    [tex] v_x= \dot{ξ}\cos{θ}-\dot{η}=\dot{ξ}\cos{θ}(1-\frac{m}{M}) [/tex]
    And the [itex]y[/itex] component is:
    [tex] v_y =\dot{ξ}\sin{θ} [/tex]
    Right?
     
  10. Dec 14, 2014 #9

    ehild

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    Yes.
     
  11. Dec 14, 2014 #10
    What about the rotational kinetic energy? What [itex] v[/itex] do you use?
     
  12. Dec 14, 2014 #11

    ehild

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    The ball rotates with angular frequency ω around the CM, it is independent from the velocity of the wedge.
     
  13. Dec 14, 2014 #12
    I believe you are overdoing things.
    Try calculating total energy at the start of the system, and then calculate at the end of the system.
    Find out an equation for velocity, use newton's third law of motion. You'll get what you are looking for.
     
  14. Dec 14, 2014 #13
    So in the end we have:

    [tex] K=\frac{1}{2}m\dot{ξ}^2(\sin^2{θ}+\cos^2{θ}(1-\frac{m}{M})^2)+\frac{2}{10}m\dot{ξ}^2 +\frac{m^2\cos^2{θ}\dot{ξ}^2}{2M}[/tex]
    [tex] V=-mgξ\sin(θ)[/tex]
    [tex] \dot{E}=m\ddot{ξ}(\sin^2{θ}+\cos^2{θ}(1-\frac{m}{M})^2)+\frac{4}{10}m\ddot{ξ}+\frac{m^2\cos^2{θ}\ddot{ξ}}{M}-mg\sin{θ}=0[/tex]
    When we solve for [itex]\ddot{ξ}[/itex] we get this expression:
    [tex]\frac{10gM^2sin{θ}}{5m^2+5m(m-M)\cos(2θ)-5mM+14M^2}[/tex]
     
  15. Dec 14, 2014 #14

    ehild

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    You can factor out ##(\dot \xi )^2## and simplify the expression for K. E=K+V =0. ##\dot E## is not what you wrote, but the equation ##
    m\ddot{ξ}(\sin^2{θ}+\cos^2{θ}(1-\frac{m}{M})^2)+\frac{4}{10}m\ddot{ξ}+\frac{m^2\cos^2{θ}\ddot{ξ}}{M}-mg\sin{θ}=0
    ##
    is correct. You madethe final expression too complicated, simplify. No need to introduce cos(2θ).
     
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