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2D Dynamics Problem

  1. Nov 24, 2013 #1
    1. The problem statement, all variables and given/known data

    A small block of mass m is sliding around the inside of an L-shaped track of radius r. The bottom of the track is frictionless; the coefficient of kinetic friction and the wall of the track is μk. The block's speed is v0 at t0=0. Find an expression for the block's speed at a time t.


    2. Relevant equations



    3. The attempt at a solution

    I'm not sure if my procedure is right. I think that the forces acting on the block on the three modified coordinate system are as follows: For the z-axis (up and down) we have the normal force and the weight of the block, and this net force ends up being zero. For the r-axis, we have that a normal force is the one that causes the centripetal acceleration (however, is it the same in magnitude as the normal force upwards?). And finally, in the tangential direction, we would have kinetic friction opposite in the direction of the block.

    As well, I'm not sure which is the velocity we need to find (tangential or centripetal or the combined velocity?). I think that the radial and tangential acceleration can give us the magnitude of the total acceleration of the block, and then find the velocity at a time t based on this "general" acceleration.

    Thank you very much in advance.
     

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  3. Nov 24, 2013 #2

    ehild

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    No, it is equal to the centripetal force.

    Correct.

    The velocity along the circular path is tangential.It is the acceleration that has both tangential and radial (centripetal) components. The normal force does not change the speed. The speed would change because of the friction. The friction is proportional to the normal force from the wall. What is the equation relating the tangential acceleration to speed?

    ehild
     
  4. Nov 25, 2013 #3
    So this "total" acceleration is the one that changes the velocity of the block? (so it would be different comparing it to centripetal velocity). Isn't the equation that relates tangential acceleration to speed:

    [itex]v=v_{o}+at[/itex]?

    So, wouldn't we just plug in the acceleration value which is the magnitude of the tangential acceleration and the centripetal acceleration? I think that we have (as the only force in the radial component is the normal force):

    [itex]a_{t}=-μ_{k}mg[/itex]
    [itex]a_{r}=g[/itex]

    The final answer at the end the book gives is:

    [itex]v=\displaystyle\frac{rv_{0}}{r+v_{0}u_{k}t}[/itex]
     
  5. Nov 25, 2013 #4

    ehild

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    The bottom of the track is frictionless. mg is a vertical force, the block slides along a horizontal track. All forces are horizontal. mg does not come in.
    The velocity is tangential, there is no "centripetal velocity".

    The tangential acceleration is related to the speed as you wrote, v=vo+att. The speed decreases because of the force of friction with the wall. at=-μN, and N is not mg, but the force the wall pushes the block inward and forces the circular motion: that is, the normal force is equal to the centripetal force.
    How is the centripetal force related to the speed?

    ehild
     
  6. Nov 25, 2013 #5
    The centripetal force would be:

    [itex]N=F_{c}=\displaystyle\frac{mv^{2}}{r}[/itex]

    And then v would be the equation we found, [itex]v=v_{0}-μNt[/itex]. So we would replace it in this equation? However, I find a second degree equation (with the mass I can't get rid of). What is the way to solve it?

    Thank you ehild for your patience.
     
  7. Nov 26, 2013 #6

    ehild

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    Your equation for the speed v(t) is not correct. The equation for the tangential acceleration is ma=mdv/dt=- μN. But N=mv2/r, plug it into the previous equation. You get mdv/dt=- μmv2/r. Simplify with m: dv/dt = -μv2/r. This is a separable differential equation for v. Can you solve it?

    ehild
     
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