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I Ball rolling down a hill?

  1. Jul 2, 2017 #1
    So I know that given a unchanging hill, and same mass between a sphere and cube, that the cube should slide down the hill faster (assuming negligible friction). This is observed through the energy "lost" by the sphere which instead of having all of its potential energy transferred towards rolling down the hill, some goes to giving the ball rotation. Now I was wondering if there is anyway to calculate of how fast a ball would roll down a hill, only given gravity.

    For example, a solid sphere (weighing 1kg with a radius of 1m) is on top of a hill that is 15m high with an slope of 30 degrees. How long will it take for the sphere to reach the bottom. Since this is an example problem an explanation is more helpful than an answer. Thanks!
     
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  3. Jul 2, 2017 #2

    russ_watters

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    You've made contradictory assumptions about friction for the sphere and cube(the sphere can't roll unless there is friction). Do you really want to do that?
     
  4. Jul 2, 2017 #3

    Orodruin

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    If there is no friction the ball will not rotate either. However, a similar problem involves homogeneous balls and cylinders rolling without slipping vs hollow balls and cylinders doing the same. They have different moments of inertia and will therefore experience different accelerations, see this image from Wikipedia:
    440px-Rolling_Racers_-_Moment_of_inertia.gif
    Edit: Note that the red sphere is the hollow sphere...
     
  5. Jul 2, 2017 #4

    BvU

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    You should realize that without friction, rolling does not originate.
    So, let's assume enough friction to have 'no slipping'. In that case potential energy from gravity is converted to kinetic energy -- a part as translational and one part as rotational.
    Do you know about angular velocity and rotational kinetic energy ?
     
  6. Jul 2, 2017 #5
    Yea I'm familiar with them, and my problem was that I could calculate the amount of energy that went towards rotational kinetic energy. PE=1/2mv^2 + 1/2Lw^2 , and this makes sense when compared to a cube that isn't rotating so would have PE=1/2mv^2 . So my question is how slower would the ball accelerate compared to the cube? I could find out how fast the cube would slide down by finding gravity's component vector pushing it and integrate, but I can't do the same for the sphere because some energy goes towards rotating the sphere. Thats where I'm stuck
     
    Last edited: Jul 2, 2017
  7. Jul 2, 2017 #6
    I'm new here and didn't really know how to word the question. It is assuming that there is static friction but that there is no kinetic friction.
     
  8. Jul 2, 2017 #7
    I'm assuming you meant to write I for moment of inertia and not L for angular momentum.

    If you first find the moment of inertia of a sphere, which you could find through easy-ish calculus or just look up, you could use conservation of energy to find the final velocity quite easily. Remember that there's a simple relation between angular and linear velocity.

    You could say there's static friction but not kinetic friction, or you could just say the kinetic friction is small enough for it to be considered negligible.
     
  9. Jul 2, 2017 #8
    Yes, you can answer it, but I guess you'd need calculus... If you don't know calculus just skip to the "which gives us:" part in the end.
    This is how I went through it:
    Through conservation of energy we know that:

    [tex]Pe = KE_T + KE_R [/tex]
    At every point, so:
    [tex]mgh = \frac{mv^2}{2} + \frac{Iω^2}{2} [/tex]
    [tex]I_{Sphere} = \frac{2}{5} mr^2[/tex]
    and
    [tex]ω=\frac{v}{r}[/tex]
    So
    [tex]mgh = \frac{mv^2}{2} + \frac{mv^2}{5}[/tex]
    giving us the velocity of the sphere at each hight covered:
    [tex]v(h)=\sqrt{\frac{10}{7}gh}[/tex]

    If it is a simple slope, you can parametrize the length [itex]s[/itex] of the ramp in terms of the hight:
    [tex]\frac{h}{Sin(θ)}=s[/tex]
    Since:
    [tex]v(h) = \frac{ds}{dt}=\frac{ds}{dh}\frac{dh}{dt} = Csc(θ)\frac{dh}{dt}[/tex]
    So:
    [tex]dt = Csc(θ)\frac{dh}{v(h)}[/tex]
    [tex]\int_{0}^{t}dt' = Csc(θ)\int_{h_f}^{h_0}\frac{dh}{v(h)}[/tex] (Switched limits of h's because its going downwards)
    [tex]\int_{0}^{t}dt' = Csc(θ)\int_{h_f}^{h_0}\frac{dh}{\sqrt{\frac{10}{7}gh}}[/tex]
    which gives us:
    [tex]t(h,θ) = Csc(θ)\sqrt{\frac{14}{5 g}}(\sqrt{h_0}-\sqrt{h_f})[/tex]

    So, in your case, we'd get:
    [tex]t(15,30°) = Csc(30°)\sqrt{\frac{14}{5\times9.81}}(\sqrt{15}-\sqrt{0})=4.13828s[/tex]
    Which is considerably slower then a block sliding down the same ramp, which would take [itex]2.4731s[/itex]

    PS: This has got to be one of the pretties solutions I've ever came up with out of the blue, kind of proud of myself :biggrin:
     
    Last edited: Jul 2, 2017
  10. Jul 3, 2017 #9
    Well, I used a different method, but we got the same result.

    We found that ##v= \sqrt { \frac {10gh} {7}}## using the same method.

    I then used two kinematic equations to solve for time:

    ##v^2_f=v^2_i+2ad⇒a=\frac {v^2_f} {2d}##
    ##v_f=v_i+at⇒t=\frac {2d} {v_f}##

    Plugging in the final velocity and then plugging in the values gave me 4.318 seconds as well.
     
  11. Jul 3, 2017 #10
    Unfortunately I don't think doing like that works, since [itex]v[/itex] is the total speed, not just the speed in the y direction, so it won't work by just plugging in the formula, since the formula only works for one dimension at a time.
    Also, it's gotta be dependent on the angle of the ramp, since going down 15m on a steep slope is much faster then in a shallow slope, and an angle of 0 would give back an infinite time (since there'd be no slope).
     
  12. Jul 3, 2017 #11
    The equation works as long as the acceleration is constant. If I know the initial velocity, the final velocity, and the displacement, and I know the acceleration is constant, I can find the acceleration; it doesn't matter what direction it's in.

    The equation I derived is dependent on distance the ball travels and height of the ramp, so therefore the angle.
     
  13. Jul 3, 2017 #12
    You're right! My bad.
    Damn, all that calculus for nothing

    and now I realize
    [tex]Csc(θ) \sqrt{\frac{14}{5g}}(\sqrt{h})[/tex]
    [tex]\frac{s}{h}\sqrt{\frac{14}{5g}}(\sqrt{h})[/tex]
    [tex]s\sqrt{\frac{14}{5gh}}[/tex]
    [tex]2s\sqrt{\frac{14}{20gh}}[/tex]
    [tex]2s\sqrt{\frac{7}{10gh}}[/tex]
    [tex]\frac{2s}{\sqrt{\frac{10gh}{7}}}[/tex]
    Exact same thing... damn...
    Well, at least it was fun deriving the whole thing :biggrin:
     
  14. Jul 3, 2017 #13
    If you take into account the work required to derive the equations I used, our methods are probably about as fast.

    I still have to often avoid using calculus because I've yet to receive a formal education on it (I have to wait until next year for that).
     
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