Derivation of parametric Equations?

  • Thread starter Mark Rice
  • Start date
  • #1
37
0

Homework Statement


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).

Homework 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

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.
 

Attachments

Answers and Replies

  • #2
PeroK
Science Advisor
Homework Helper
Insights Author
Gold Member
13,761
6,253

Homework Statement


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).

Homework 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

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.
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
37
0
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.
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
37
0
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?
I've just realised that is not even close to being right...
 
  • #5
37
0
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.
I understand what you're saying to do, but I'm not sure how I acutally work that out though
 
  • #6
37
0
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.
is the parametric equation y(t)=1/2π t sin (π/6)
 
  • #7
PeroK
Science Advisor
Homework Helper
Insights Author
Gold Member
13,761
6,253
I understand what you're saying to do, but I'm not sure how I acutally work that out though
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
PeroK
Science Advisor
Homework Helper
Insights Author
Gold Member
13,761
6,253
is the parametric equation y(t)=1/2π t sin (π/6)
The position of a point on the wheel is going to be a function of ω, θ and ##t##.
 
  • #9
37
0
The position of a point on the wheel is going to be a function of ω, θ and ##t##.
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
37
0
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
The position of a point on the wheel is going to be a function of ω, θ and ##t##.
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
PeroK
Science Advisor
Homework Helper
Insights Author
Gold Member
13,761
6,253
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
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
37
0
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.
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 :)
 

Related Threads on Derivation of parametric Equations?

  • Last Post
Replies
2
Views
2K
  • Last Post
Replies
4
Views
1K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
5
Views
1K
  • Last Post
Replies
3
Views
2K
Replies
6
Views
8K
  • Last Post
Replies
9
Views
2K
  • Last Post
Replies
4
Views
4K
  • Last Post
Replies
4
Views
1K
Top