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Here is the sketch:

Sketch for the problem

The sketch is supposed to be side-view of the path of the object.

The following values are known:

  • $r$ - radius of the circle that describes the path AB of the object
  • $a$ - angle that characterizes the part of a circle that describes the path AB of the point
  • $m$ - mass of the point
  • $V_0$ - velocity

What I need to find out:

  1. Equation of motion for AB
  2. Equation of motion for BC
  3. velocity at B
  4. The distance DC

The dashed line is the object's trajectory after it leaves AB. $N$ is the normal force, $T$ is friction and $g$ is the gravitational acceleration.

I was able to solve this problem partially when AB is a straight line and $a$ represents the angle between AB and AD. So far I could come up with only this:

$m x'' = -T-mg \sin(?)$ <- in place of the question mark I would need the angle between AB and AD

$m y'' = N-mg \cos(?)$

$N = mg \cos(?)$

$T = \mu N = \mu mg \cos(?)$

$x'' = -g(\mu \cos(?) + \sin(?))$

$x' = -gt(\mu \cos(?) + \sin(?)) + c_1$

$x = g\frac{-t^2}{2}\left[ \mu \cos(?) + \sin(?) \right] + c_1 t + c_2$

where $\mu$ is the coefficient of friction. $x$ and $y$ are the coordinates with respect (both functions of time).

How do I deal with the fact the ramp is no longer a straight line but a curved line? Thank you very much for your help.

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closed as off-topic by Kyle Kanos, BMS, garyp, John Rennie, Brandon Enright Jun 26 '14 at 16:03

This question appears to be off-topic. The users who voted to close gave this specific reason:

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  • $\begingroup$ More information required! Is this a graph? What happens at B? What is the dashed curve? How does the body even get to D? $\endgroup$ – Rob Jeffries Jun 26 '14 at 15:51
  • $\begingroup$ I agree with Rob, something is off either in the description of what we are trying to solve for or in the plot itself. What I think you mean is the equation of motion for AB and then BC. $\endgroup$ – Bryson S. Jun 26 '14 at 16:05
  • $\begingroup$ I'm sorry, I mixed up the letters. It is as you said. I corrected it. $\endgroup$ – Tamás Jun 26 '14 at 16:07
  • $\begingroup$ I've taken the liberty of LaTeXifying your work, but I may have mis-interpreted your intent somewhere, so this should be checked. We have MathJax running on the site which means that math can be written in a $\LaTeX$-math-mode-alike language between single $ for in-line equations and double $$ for block set. I.e. $ax^2 + bx + c = 0$ is typeset as $ax^2 + bx + c = 0$. $\endgroup$ – dmckee Jun 27 '14 at 14:32
  • $\begingroup$ I see. I'm gonna enclose math expressions in $ from now on. Thank you for editing btw $\endgroup$ – Tamás Jun 27 '14 at 14:37
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HINT: From conservation of energy, you can find the velocity at the release point; after that it will follow a projectile motion. You can thus write equation of trajectory from the projectile motion.

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    $\begingroup$ I think you mean C.o.E. can be used to find the velocity at the release point, not the highest point. Based on the figure, the particle is experiencing projectile motion long before it reaches it's highest point $(\alpha>0)$. $\endgroup$ – Bryson S. Jun 26 '14 at 16:14
  • $\begingroup$ @BrysonS. Yes... $\endgroup$ – RE60K Jun 28 '14 at 13:45

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