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Block moving on an inclined plane

  1. May 11, 2014 #1

    utkarshakash

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    1. The problem statement, all variables and given/known data
    A block is placed on a plane inclined at an angle θ. The coefficient of friction between the block and the plane is µ = tan θ. The block is given a kick so that it initially moves with speed V horizontally along the plane (that is, in the direction perpendicular to the direction pointing straight down the plane). What is the speed of the block after a very long time?


    3. The attempt at a solution

    The force of friction exactly balances the gravitational force along the incline. So there will be no acceleration along the inclined plane and the trajectory of the block must be a straight line and it continues to move in its original direction of motion. The frictional force will act opposite to its velocity and after a long time the block should stop. So the final velocity should be 0. But this is not the correct answer. I can't figure out what's wrong with my reasoning.
     

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  3. May 11, 2014 #2
    You are ignoring the fact that the force of kinetic friction is always aligned with the velocity. So initially it is purely horizontal and does not oppose the force of gravity that pulls the block sideways, so the velocity vector gets a sideways component, so does the force of friction, etc.
     
  4. May 12, 2014 #3

    utkarshakash

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    So how should I proceed?
     
  5. May 12, 2014 #4
    The first step here would be to write down the components of the net force acting on the particle in the plane of the incline.
     
  6. May 12, 2014 #5

    utkarshakash

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    The net force is mgsinα down the incline and frictional force = mgsinα opposite the direction of initial velocity
     
  7. May 12, 2014 #6
    It would be useful to have that written as differential equations.
     
  8. May 12, 2014 #7

    ehild

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    The force of friction is opposite to the instantaneous velocity.

    ehild
     
  9. May 12, 2014 #8

    haruspex

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  10. May 12, 2014 #9
    There is a very simple integral that can be reached via a mathematical device, or physical intuition.
     
  11. May 12, 2014 #10

    ehild

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    You do not need to solve the differential equation. The problem asks the speed after very long time.
    If you write the differential equations for the velocity components along the slope(vx) and in the horizontal direction (vy), and also the derivative of the speed v, you get a very simple relation between dv/dt and dvx/dt. Also, it is easy to figure out the long-term behaviour of vy.

    But I would like to see the solution for v(t)...

    ehild
     
  12. May 12, 2014 #11
    Agreed. I would like to see the complete solution as well.
     
  13. May 13, 2014 #12

    haruspex

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    Ok.
    The forces are mg sin(θ) down the plane, and likewise in the direction opposite to the velocity.
    In (s, ψ) coordinates, with ψ as the angle of the velocity to the horizontal in the plane,
    ##\ddot s = (\sin(\psi)-1)g \sin(\theta)##
    ##\dot s \dot \psi = g \sin(\theta) \cos(\psi)##
    Writing ##\dot s = g \sin(\theta) u##, this simplifies to
    ##\dot u = (\sin(\psi)-1)##
    ##u \dot \psi = \cos(\psi)##
    Eliminating u we get:
    ##\dot\psi^2(1-2 \sin(\psi)) = \ddot \psi \cos(\psi)##, with initial conditions ##\psi(0) = 0, \dot \psi(0) = \frac 1{u(0)} = \frac v{g \sin(\theta)}##
    Rewriting that as ##\dot\psi(\sec(\psi)-2 \tan(\psi)) = \frac{\ddot \psi }{\dot \psi }## we can integrate to obtain
    ##\dot \psi = A \cos(\psi)(1+\sin(\psi))##, ##u = \frac 1{A(1+\sin(\psi))}##
    The initial conditions give ##A = \frac{g \sin(\theta)}{v}##, so ##\dot s = \frac v {(1+\sin(\psi))}##
    We want ##\mathop{\lim}_{t\rightarrow \infty}\dot s = \mathop{\lim}_{\psi\rightarrow \frac{\pi}{2}}\dot s = \frac v2##
    (What's a better way to write limits in LaTex?)
    You could also solve to find t as a function of ψ.
     
  14. May 13, 2014 #13

    ehild

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    $$\mathop{\lim}_{t\rightarrow \infty}\dot s = \mathop{\lim}_{\psi\rightarrow \frac{\pi}{2}}\dot s = \frac v2$$

    Use $$ instead of ##, but then it becomes a new line.

    ehild
     
  15. May 13, 2014 #14
    Nicely done, haruspex. For limits, there is a "limits" modifier that forces indices above and below the current symbol: ##\lim\limits^{up}_{down} \int\limits_{from}^{to}\sum\limits_{begin}^{end}##.
     
  16. May 13, 2014 #15

    utkarshakash

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    Can you please explain what have you done here? It's difficult for me to follow your approach. What does (s,ψ) coordinates mean here?
     
  17. May 13, 2014 #16
    You do not need to follow that. Use the approach mentioned by ehild.
     
  18. May 13, 2014 #17

    TSny

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  19. May 13, 2014 #18

    utkarshakash

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  20. May 13, 2014 #19

    haruspex

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    As ehild and voko point out, there's a much easier way to obtain ##\dot s = \frac{v}{1+\sin(\psi)}##. If not, I would have given hints, not posted my calculus solution in detail.
    In (s,ψ) coordinates, s is the distance along the path from some fixed origin, and ψ is the angle the path makes to some fixed direction at that point. (I chose that to be the initial direction of motion, with positive ψ being down the plane.) The link I posted shows how to convert to and from other systems.
    It gives much simpler differential equations in some cases, like this one. The acceleration along the path, ##\ddot s ##, comes from the frictional force plus the sin(ψ) component of the downplane gravitational force:
    ##\ddot s = (\sin(\psi)-1)g \sin(\theta)##
    The acceleration normal to the path, ##\dot s \dot \psi##, comes from the cos(ψ) component of the downplane gravitational force:
    ##\dot s \dot \psi = g \sin(\theta) \cos(\psi)##
    Since g sin(θ) occurs as a common factor, it can be eliminated by incorporating it into s.
     
  21. May 13, 2014 #20

    ehild

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    Cartesian coordinates might be used, too, and that approach might be more familiar to the OP.

    The x axis is horizontal, directed along the initial velocity. The y axis is parallel with the slope, pointing downward. See picture.

    The downward force is Fg=mgsinθ. The magnitude of the force of friction is also mgsinθ. The force of friction is opposite to the velocity v. It can be written as the negative of the magnitude multiplied by the unit vector in the direction of the velocity $$\vec F_f=-mgsinθ \frac{\vec v}{v}$$
    If vx, vy are the components of the velocity and v is the speed ## v= \sqrt{v_x^2+v_y^2}## the horizontal component of the friction is ##-mg\sinθ \frac{v_x}{v}## and the vertical component is ##-mg\sinθ \frac{v_y}{v}##.

    Write out the x and y components of acceleration, and also the time derivative of the speed v. Use the notation A=gsinθ.

    $$\dot v_x=-A\frac{v_x}{v}$$
    $$\dot v_y=A(1-\frac{v_y}{v})$$

    $$\dot v=\frac{v_x \dot v_x +v_y \dot v_y}{v}$$

    Substitute the first two equations for vx, vy into the third one. You will see that the time derivative of vy is equal and opposite to that of v. Integrate.

    It can be proved mathematically, but it is clear that the horizontal component of the velocity tends to zero with time, so the speed becomes equal to the vertical velocity component v→vy.

    ehild
     

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