# Mechanics - Curvilinear Motion x/y coordinates

1. Aug 15, 2009

### aolse9

1. The problem statement, all variables and given/known data
The quality assurance procedure for ball bearings has them coming down a shute and bouncing through a hole in a plate as shown. Acceptable balls pass through the hole and impact the inclined slope at Q. Find HW

http://hdimage.org/images/5plyw1jdci01pyz641pb.jpg [Broken]

2. Relevant equations

3. The attempt at a solution
I have been staring at this question for the past day, trying to think of how to approach it. However it has been to no avail. I am stumped, and I would really appreciate some help to steer me in the right direction. Cheers.

Last edited by a moderator: May 4, 2017
2. Aug 15, 2009

### kuruman

Note that there is no vertical component of the velocity when the ball goes through the hole. What is its speed then? Find an expression (y = f(x)) for the parabolic trajectory after the ball passes through the hole. Find an equation for the straight line at 48.0o. Find the point of intersection of the parabola and the straight line.

3. Aug 17, 2009

### aolse9

Yeah, so it will have a velocity of 0ms-1 at the vertex. I am having trouble finding the expression for the parabolic trajectory. Do I let P be the origin and define the equation of the parabola and straight line using the coordinates given? Also how do you define the a, b and c values in the general form of a parabola, y = ax2 + bx + c?

4. Aug 17, 2009

### djeitnstine

Use newtons basic equations of motion. When you derive the equation for parabolic motion neglect all external forces other than gravity.

Acceleration vector looks like <0,-g> (Using Up as positive y of course)

5. Aug 17, 2009

### kuruman

Yes, use P as the origin. To get the parabola, write the two kinematic equations for projectile motion giving x(t) and y(t). Solve the first equation for t and substitute in the second. This eliminates t and gives you an equation for y in terms of x. Once you have that, you can easily identify a, b, and c.

6. Aug 18, 2009

### kuruman

It was written in haste as I initially thought that the acceleration of gravity must be part of all this. It does not.

The origin should be directly underneath the hole where the inclined plane starts. This makes writing the equation for the straight line trivial. For the parabola, note that, if the incline were not there, y is zero at two points, x = -L and x = +L (L = 121 mm). These are the roots of a quadratic. So the parabola can be written as

y(x) = C (x - L)(x + L)

All you need to do is find C such that y(0) = 143 mm

7. Aug 19, 2009

### djeitnstine

Ok I did some pen and paper calculations.

Finding the equation of the parabola is actually quite simple. Place the origin at P as kuruman suggested.

Then you take the general equation of a parabola $$y=ax^2 + bx + c$$

Since we're at the origin c = 0 right away.

Then we take the derivative y' $$y' = 2ax +b$$

We have 1 crucial piece of information. y(121) = 143 i.e.

$$143=a121^2 + b121$$ and since this is the location of the max from pre calc we know

y' = 0
$$0 = (2)(121)a +b$$

we now have 2 equations and 2 unknowns.

to find $$h_w$$, we now need to find an equation for that line.

I would suggest start by placing a new origin at x=121mm and use some geometry to find the line at 48 degrees that intersects the parabola. With the new origin there do not forget to make c=143

8. Aug 19, 2009

### djeitnstine

to find the slope of the line here is a hint:

The slope is related to basic trigonometry.

9. Aug 19, 2009

### kyiydnlm

From P to hole:
t = (2*0.143/9.81)^0.5
vx = 0.121 / t
vx is a constant if air resistance does not present.

From hole to Q:
Hw = 0.143 - 0.5*9.81k -------------- (1)
Qx = vx*k --------------------------- (2)
Hw = Qx*tan(48) ----------------------(3)
(k is elapsed time from hole to Q)

three equations, three unknown.
you can make one equation for (1), (2) and (3) if you want.