Derivation of parametric Equations?

1. Oct 30, 2014

Mark Rice

1. The problem statement, all variables and given/known data
Hi, so confused abou this question that I probably haven't even posted it in the correct section.Here's the question.

A wheel of radius ,r, is situated at the top of a ramp having an angle θ = π/6 rad. At t= 0 the wheel is at rest with its centre at coordinates (0,r) and then it starts to rotate clockwise, without slipping, in the positive x direction with constant angular velocity ω.

Find the parametric equations of the x and y coordinates of the point p (green dot on graph) with respect to the t >= 0, assuming the initial co-ordinates are at (0, 2r).

2. Relevant equations
I think possibly ω=ω+at but I don't even know what is going on. If it helps in anyway I know the circle has centre (0, r) radius r so the equation would be x^2 + (y-r)^2 = r^2

3. The attempt at a solution
The reason I'm so confused is that I'm sure they haven't taught us this work yet. I think I do it by working out the parabola motion of p then split that into equations for the horizontal and vertical motion, but literally no idea how I would even go about doing that let alone writting it down mathematically. ANY help would be great. Thanks.

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2. Oct 30, 2014

PeroK

Why not break the problem down: first work out the equation for the centre of the wheel (x(t), y(t)); then work out the position of the point p relative to that.

3. Oct 30, 2014

Mark Rice

I honestly don't even know how to do that. I think the point might posibly y(t)=2π/t but am pretty sure that's not right?

4. Oct 30, 2014

Mark Rice

I've just realised that is not even close to being right...

5. Oct 30, 2014

Mark Rice

I understand what you're saying to do, but I'm not sure how I acutally work that out though

6. Oct 30, 2014

Mark Rice

is the parametric equation y(t)=1/2π t sin (π/6)

7. Oct 30, 2014

PeroK

Well, you have to decide whether to attempt this problem. Could you work out the parametric equations for a point on a wheel moving along horizontally? That would be a simpler problem.

8. Oct 30, 2014

PeroK

The position of a point on the wheel is going to be a function of ω, θ and $t$.

9. Oct 30, 2014

Mark Rice

If this isn't close I have no hope, is the function x=rcos(ωt+θ) and y=rsin(ωt+θ) ? Sorry about being so bad at this

10. Oct 30, 2014

Mark Rice

Sorry just realised that they are wrong I think they should be x=vt+rsin(ωt+θ) and y=y0+rcos(ωt+θ) but this is for horizontal movement so how do I work it out for on a slope?

11. Oct 30, 2014

PeroK

You'll not get anywhere just guessing! Try the problem along the flat. First, you have to work out how fast the wheel is moving. Second, you'll have to work out where point p is relative to the centre of the wheel.

If you can do this, then you can extend the solution to be down the slope.

12. Oct 30, 2014

Mark Rice

Really sorry but I literally have no idea what is going on in this question so am going to just leave it. Greatly appreciate all the help you gave but I'm way too confused. Thanks anyway :)