Mechanics - Acceleration on an elliptical path

[Kaevan807]

This is a mechanics question which has come up a number of times in end of year exams in my college, I hope this is the right forum to post it in.

Homework Statement

A particle travels around an elliptical path with a constant speed of 1 m/s. If the equation
of the ellipse is (x^2)/3 + y^2 = 1
Determine the maximum magnitude of the acceleration.

A - (√3)/2
B - 3/2
C - 1/2
D - 0
E - 1

Homework Equations

If the equation = (x^2)/(a^2) + (y^2)/(b^2) = 1
Then a = (v^2)/ρ, where ρ is the radius of curvature and is equal to (a^2)/b

The Attempt at a Solution

Using the above equations gives:
ρ = 3/1 = 3
a = 1/3 which is not an answer provided?


I'm unsure whether the answer should just be zero? As there is no tangential acceleration? Or do I have the wrong formula for calculating the centripetal acceleration?

Any help or explanation would be great :)

 

[TSny]

Then a = (v^2)/ρ, where ρ is the radius of curvature and is equal to (a^2)/b

Is this the maximum or minimum radius of curvature of the ellipse?

 

[Kaevan807]

This is the maximum radius of curvature.

 

[TSny]

This is the maximum radius of curvature.

But you want to find the maximum acceleration.

 

[Kaevan807]

But you want to find the maximum acceleration.

Yes but isn't the maximum centripetal acceleration when the particle is traveling at the point of largest radius?

 

[cepheid]

Yes but isn't the maximum centripetal acceleration when the particle is traveling at the point of largest radius?

No...

 

[TSny]

Remember:

... a = (v^2)/ρ, where ρ is the radius of curvature ...

 

[Kaevan807]

Had an entire post written and then realized that it's the smaller radius but furthest from centre of the ellipse.

Ok so taking that into account would the answer be 1?

Minimum radius = 1

a = (v^2)/r = 1/1 = 1?Edit
Thank you, can't believe I didn't get that >.< That's a great help.

 

[TSny]

Ok so taking that into account would the answer be 1?

Minimum radius = 1
a = (v^2)/r = 1/1 = 1?

No, the minimum radius of curvature of the ellipse is not 1. a²/b is the curvature of the ellipse at a point on the y-axis. This happens to be the maximum radius of curvature for your ellipse because the ellipse is elongated along the x-axis. Think about where the minimum curvature will occur and see if you can guess how the formula a²/b would change to get the minimum curvature.

 

[Kaevan807]

Dang just when I thought I had it.

Well I'm guessing the equation will change to b²/a?

Although that gives me 1/√3 which when subbed into a doesn't result in an answer?


Edit:

I was going under the assumption that a and b are the two radii? So whichever is smaller I just sub directly into a=v²/r, this gives me an answer for all 3 years this question has come up

 

[TSny]

Dang just when I thought I had it.

Well I'm guessing the equation will change to b²/a?

Although that gives me 1/√3 which when subbed into a doesn't result in an answer?

I think 1/√3 is the right answer even though it's not listed as one of the choices.

I was going under the assumption that a and b are the two radii? So whichever is smaller I just sub directly into a=v²/r, this gives me an answer for all 3 years this question has come up

I hope whoever wrote the question wasn't assuming that the smallest radius of curvature would be the smaller of a and b!

 

[Kaevan807]

Written by a mechanics lecturer in my college but unfortunately I haven't had a reply to any email I've sent him. The questions the other two years were:
1. A particle travels around an elliptical path with a constant speed of 1 m/s. If the equation
of the ellipse is 3x² + 2y² = 2
Determine the maximum magnitude of the acceleration.

A - (√3)/2
B - √3/√2 (Answer using the radius being the smaller of a and b)
C - 1/2
D - 0
E - 1

2. A particle travels around an elliptical path with a constant speed of 1 m/s. If the equation of the ellipse is x²/2 + y² = 1
Determine the maximum magnitude of the acceleration

A - (√3)/2
B - 3/2
C - 1/2
D - 0
E - 1 (Answer using the radius being the smaller of a and b)


As you can see neither of these has an answer which would work using the b²/a formula

 

[TSny]

For #1, a is smaller than b. So, the smallest radius of curvature is a²/b = 2/3. Max acceleration is then 3/2.

For #2, b is smaller than a. So, the smallest radius of curvature is b²/a = 1/√2. Max acceleration is √2.

As you noted, neither of these corresponds to any of the selections. Hope I'm not overlooking something.

 

[Kaevan807]

Unless he meant the magnitude of the tangential acceleration? Which would be zero each time? But he doesn't say that