Ball rolling down smooth curve

[Mardoxx]

Ball sliding down smooth curve

Homework Statement

A mass of 2 kg is dropped vertically into a frictionless slide located in the x-y plane. The mass enters with zero velocity at (-5,5) and exits traveling horizontally at (0,0).

Assuming the slide to be perfectly circular in shape construct a model of the forces acting on the mass and hence simulate its motion.

Homework Equations

Unknown.

The Attempt at a Solution

How is it done?

This is the question written so it's easier to understand...

Quarter-Circle radius 5 starting at (-5,5), ending at (0,0), centre at (0,5).
Ball mass 2.1kg is let go vertically at (-5,5) and left to slide down the slope.
The ball can be assumed to be a sliding point mass.

So,
s_0 = (-5,5)
v_0 = (0,0)
a_0 = (0,-9.8)

v_f = (sqrt(2*9.81*5),0) = (9.90, 0)
s_f = (0,0)

But, how do you model it?

 

[lanedance]

the first thing to do is describe the force on the ball at each point on the circle

to do this, you should parametise equation of the cricle in terms of a single variable, angle probably being a good choice

 

[HallsofIvy]

By the way, you titled this "ball rolling down smooth curve". It is an important simplification in this problem that the slide is "frictionless" and so the ball does NOT roll! That means that you can consider the ball as a single point (its center of mass) and find the motion of that.

 

[Mardoxx]

ok, so the parametric equation for a circle radius 5 origin 0,5 is
[tex]y = 5cos( \theta )+5; x = 5sin( \theta )[/tex]

or in Cartesian form

[tex](x)^2 + (y-5)^2 = 25[/tex]

would the velocity at any given y be
[tex]v^2 = 2 g (y-5)[/tex]
or would that just be the y component of the velocity?

 

[lanedance]

Halls makes a good point & as you haven't siad anything about ball mooment of inertia or radius, i will just assume it is a sliding point mass

first, the circle centre is actually (5,0) as i read it

if you want to use energy balance, the difference between total kinetic energy & gravitational potential will be conserved, so that should help you decide what v represents in the equation you gave

 

[Mardoxx]

centre is 0,5

but yeah, it's a sliding point mass.
the difference between the two is the same through out? how? becuase it would have 0 GPE at 0,0 :S

 

[lanedance]

sorry i meant sum

 

[Mardoxx]

ok, so v is tangential velocity.
so how do I go from tangential velocity to position (x,y) at time t?

 

[lanedance]

so kinetic energy is based on the magnitude of velocity

so for a given point on the curve you now know velocity magnitude & direction...

 

[Mardoxx]

I had another go at it, this is as far as I got:
Thanks so far btw! :)



A mass of 2 kg is dropped vertically into a frictionless slide located in the x-y plane. The mass enters with zero velocity at (-5,5) and exits traveling horizontally at (0,0).

Assuming the slide to be perfectly circular in shape construct a model of the forces acting on the mass and hence simulate its motion.


See http://imgur.com/EutWV.png

Parametric form of the path:
[tex]\begin{align*}x & = 5 Sin[\theta] \\
y & = 5 Cos[\theta]+5
\end{align*}[/tex]
Since [tex]\phi = \theta - \pi[/tex] and [tex]\phi[/tex] is the angle we're interested in, the equation can be rewritten as:

[tex]\begin{align*}x & = 5 Sin[\phi + \pi] \\
y & = 5 Cos[\phi + \pi]+5
\end{align*}[/tex]

Since the curve is smooth and it is assumed there is no air resistance we say that all GPE is converted to KE. That is,
[tex]m g h = \frac{1}{2}m v_t^2[/tex]
So we can calculate the tangential velocity [tex]v_t[/tex] at any [tex]h[/tex].
[tex]v_t = \sqrt{2gh}[/tex]
Since, in this case, [tex]h[/tex] is just our [tex]y[/tex] co-ordinate we can get [tex]v_t[/tex] in terms of [tex]\phi[/tex].
[tex]\begin{align*}v_t & = \sqrt{2gh} \\
& = \sqrt{2gy} \\
& = \sqrt{2g(5 Cos[\phi + \pi]+5)} \\
v_t & = \sqrt{10g(1+Cos[\phi + \pi])}
\end{align*}[/tex]
Since [tex]v_t = r \dot{\phi} = r\omega[/tex], where [tex]r[/tex] is the radius of the circle's path, then
[tex]\begin{align*}5 \dot{\phi} & = \sqrt{10g(1+Cos[\phi + \pi])} \\
\dot{\phi} & = \frac{1}{5}\sqrt{10g(1+Cos[\phi + \pi])}
\end{align*}[/tex]
So the angular velocity, [tex]\omega[/tex], at any given [tex]\phi[/tex] is;
[tex]\frac{d\phi}{dt} & = \frac{1}{5}\sqrt{10g(1+Cos[\phi + \pi])}[/tex]

 

[Mardoxx]

I've got this now

http://mathbin.net/45802

 

[lanedance]

no worries, method looks ok, though i haven't checked working,

angles are a little confusing i would just start with angle from horizontal, moving in direction of the motion then [itex] (x,y) = 5(-cos(\phi), 1-sin(\phi)) [/itex] but dond;t let this defintion confuse things.

one other point, the velocity of any particle constrained to a path will always be tangential to that path (you used transverse somehere as well which doesn't quite make sense)

 

[lanedance]

also note, that as you used energy balance & angluar acceleration, you haven;t actually solved for the forces anywhere if that is required...

 

[Mardoxx]

I've done it :D

Using Runge-Kutta method..

but now I need to work out vectors for the forces...

 

[lanedance]

should be just trig problem