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Block Sliding Down Circular Ramp (w/ Friction)

  1. Dec 21, 2011 #1
    A block of mass m is positioned at the top of a quarter-circle ramp with radius R and friction coefficient μ. To find the velocity v at the bottom, I did the force summation equations in the n- and t-directions and came up with:

    n: N - mgsinθ = mv^2/R
    t: mgcosθ - μN = ma

    I then solve for N in the n-direction equation, then subsitute into the t-direction equation. Solving for a, I get:

    a = g(cosθ - μsinθ) - μv^2/R

    I realize that I need to use this in the kinematic equation: vdv = ads, or in this case, vdv = aRdθ. Doing this, I get:

    vdv = (Rg(cosθ - μsinθ) - μv^2)dθ

    This can supposedly be rearranged into an easily solvable differential equation in the form of dy/dx + f(x)y = g(x), but I'm not seeing it. I'm assuming that v = y and θ = x, which would make the equation dv/dθ + f(θ)y = g(θ).

    What am I missing?
     
  2. jcsd
  3. Dec 21, 2011 #2
    Hi Bakeman
    Welcome to PF !!

    Well i haven't been in touch with physics for i long time so i might be wrong but here's a thought ...

    The method you are approaching with is based on assumption that the block will remain stick to the surface of quarter circle during complete motion ... but i don't think that is true ...

    It will leave its contact as soon as [itex]\frac{m v^2}{R} > mgsin\theta - N[/itex]
     
  4. Dec 22, 2011 #3
    The ramp is probably oriented so that the flat point rests on the ground not in the air so the block will stay in contact with it through the whole motion.
    I would use the work energy theorem to solve this one; kinetic energy = work done by gravity - work done by friction.
     
  5. Dec 22, 2011 #4
    I've been going back through some of my old college texts and found this problem that I wanted to try to solve. The problem comes before any discussion of work, kinetic energy or potential energy. The problem itself provides tips that basically tell you how to solve it: write the n and t equations, eliminate N, and substitute into vdv = ads. At that point, you should be able to rearrange into the form dy/dx + f(x)y = g(x). That's where I'm getting stuck. I know I can use the work energy theorem, but I'd like to solve it using the book's direction.
     
    Last edited: Dec 22, 2011
  6. Dec 22, 2011 #5
    I attached a picture of the situation. It is assumed that the block always stays in contact with the ramp
     

    Attached Files:

  7. Dec 22, 2011 #6
    That doesn't matter ... even if it came before the work energy discussion ... a physics problem can be solved in so many ways ...and i like the idea of JHamm

    Oops ... i thought this is the case:
    https://www.physicsforums.com/attachment.php?attachmentid=42144&stc=1&d=1324571374

    and "kinetic energy = work done by gravity - work done by friction" will perfectly.

    and keep in mind that gravity is conservative and friction is non conservative ...
     

    Attached Files:

  8. Dec 22, 2011 #7
    I understand, but I would still like to solve it according to the direction given in the problem. I'm not as interested in the answer as I am in the method that is being described in this case.
     
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