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Interesting snowball down slope; increasing mass

  1. Jan 1, 2013 #1
    1. The problem statement, all variables and given/known data

    a)A snowball, initially of mass m0, slides down a slope of incline angle ∅. As it moves, mass of additional snow Δm = αx. Write down the differential equation of translational motion for the snowball, ignoring rotation and friction.

    b)If the snowball starts from rest, derive an expression for its velocity as a function of time.

    c)A spherical ball of mass m and radius r rolls down a slope at an angle ∅. Find the frictional force required between the slope and the ball if there is no slipping.



    3. The attempt at a solution

    (a)
    By using conservation of momentum at t and t+Δt, where taking the limit Δt→0,
    Δp/Δt = mg sin ∅

    I arrive at:

    sdoimu.png


    (b)
    I change the differential equation to dv/dm first, then use a substitution.
    1z4khoy.png

    (c)
    For part (c), do I assume that it is pure rolling?
     
  2. jcsd
  3. Jan 1, 2013 #2

    mfb

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    (c): I think so.

    I think the result in (b) is not correct: For large x, v~x. Without picking up additional mass (i. e. a simple slope and a sliding mass), v~sqrt(x). In other words, your additional snow would speed up acceleration, which looks wrong.
     
  4. Jan 1, 2013 #3
    I'm not sure what you mean...besides I can't find anything wrong with my working in (b)..
     
  5. Jan 1, 2013 #4

    mfb

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    Consider a mass sliding on the same slope, without snow: It will follow the equation
    $$v^2=2g x \sin(\phi)$$
    Your snowball picks up additional mass, which slows acceleration. But your equation gives larger velocity values.

    Another way to see the problem:
    A lower bound on your velocity value is ##v>\sqrt{2g \sin(\phi) \frac{\alpha}{m_0}} x##
    As everything in the square root is constant and x increases without limit, your snowball will eventually accelerate quicker than g, which is impossible.

    I don't know where the error is (probably somewhere in the substitution part), but there has to be an error somewhere.
     
  6. Jan 1, 2013 #5
    Yeah, but the problem is I can't find anything suspicious/wrong about my working!
     
  7. Jan 1, 2013 #6
    Check your 3rd-5th lines of part b. I don't understand how you got such simple expressions for the integrals given you had sums and differing powers of integrated variables.

    mfb also appears to be wrong. Your velocity must increase due to a simple application of energy conservation (assuming the mass is added slowly enough that the ball keeps moving). So your velocity can certainly increase without bound, and what he posted has no bearing on acceleration becoming larger than g-- he posted an equation for velocity anyway! He seems to assume v increasing with x require it to increase past g with time, but clearly it could also increase with distance at slower rates in time than g.

    Still it's that integral that looks fishy to me.
     
    Last edited: Jan 1, 2013
  8. Jan 1, 2013 #7

    mfb

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    I should have made the acceleration argument more explicit:

    ##v = c x## can be written as ##\frac{dx}{dt} = cx##, and the solution is an exponential function: ##x(t)=d e^{cx}##. The acceleration is the second derivative, therefore ##v(t)=c^2d e^{cx}##. As c>0 and d>0, this grows exponentially without limit.
     
  9. Jan 1, 2013 #8

    TSny

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    Looks like the two u's that are circled in the figure were dropped.

    You can write the term ##\alpha v \frac{dv}{dm}## as ##\frac{\alpha}{2} \frac{dv^2}{dm}##. Then let ##u = v^2## and get a differiential equation for ##u## as a function of ##m## that can be solved. This can be used to give you ##v## as a function of ##x##. I haven't been able to get an explicit result for ##v## as a function of time.
     

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  10. Jan 1, 2013 #9
    Take a look at this for part b) (which also answers a) ). I combine the equations:

    [itex]dm=adx\Rightarrow \dot{m}=av[/itex] and [itex]\frac{d}{dt}\left( mv \right)=g\sin \varphi [/itex]

    to get the separable differential equation:

    [itex]m\dot{v}+a{{v}^{2}}=g\sin \varphi [/itex]

    which is solved as:

    [tex]\begin{align}
    & \frac{m\dot{v}}{g\sin \varphi -a{{v}^{2}}}=1\Rightarrow \int{\frac{mdv}{g\sin \varphi -a{{v}^{2}}}}=\int{dt}\Rightarrow \frac{m}{\sqrt{ag\sin \varphi }}\operatorname{arc}\tanh \left( v\sqrt{\frac{a}{g\sin \varphi }} \right)=t+c\Rightarrow \\
    & \Rightarrow v\left( t \right)=\sqrt{\frac{g\sin \varphi }{a}}\tanh \left[ \frac{\sqrt{ag\sin \varphi }}{m}\left( t+c \right) \right] \\
    \end{align}[/tex]
     
  11. Jan 1, 2013 #10

    TSny

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    But m is not a constant in the integral.
     
  12. Jan 1, 2013 #11
    oops.:redface: I forgot about it.... I'll see if I can do something else
     
  13. Jan 3, 2013 #12
    Any other ways to do solve the differential equation??
     
  14. Jan 3, 2013 #13

    mfb

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    I tested the direct approach and got ugly differential equations.

    Here is another idea:
    Replace ##g\sin(\phi) \to g## to simplify equations (we can replace it back again afterwards).
    Define x as ##\frac{m_0}{\alpha}## above the starting point, this gives ##m(x)=\alpha x## and ##\frac{dm}{dt}= \alpha v##

    $$\frac{dv}{dt} + \frac{v}{m} \frac{dm}{dt} = g$$
    Substitute m and its derivative, use ##\frac{dv}{dt}=\frac{dv}{dx}\frac{dx}{dt}=v \frac{dv}{dx}##:
    $$v\frac{dv}{dx} + \frac{v^2}{x} = g$$
    Simplify:
    $$\dot{v} + \frac{v}{x} = \frac{g}{v}$$
    Use WolframAlpha (lazy version)
    $$v=\frac{\sqrt{c+2gx^3}}{\sqrt{3}x}$$
    Where ##v(\frac{m_0}{\alpha})=0## can be used to fix c.

    Unfortunately, this is v as function of x, so we need more steps to switch to a function of time.
     
  15. Jan 4, 2013 #14
    I'm not sure how you arrived at the last step, as

    m = M0 + αx

    and

    dm/dt = αv

    sdoimu.png
     
  16. Jan 4, 2013 #15

    mfb

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    I used a different definition of x, to remove m0 (it is included in the starting position of the snowball).
     
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