Elliptical motion about the origin

  • Thread starter Thread starter Identify
  • Start date Start date
  • Tags Tags
    Motion Origin
Click For Summary
SUMMARY

The discussion focuses on determining the period of a ball of mass m following an elliptical path defined by the equation r(t)=bcos(ωt)e(x)+2bsin(ωt)e(y). The consensus is that the period T is given by T=2π/ω, as both the cosine and sine components have a period of 2π. Additionally, participants discuss the simplification of the distance from the origin using the identity sin²(u) + cos²(u) = 1, emphasizing the importance of recognizing coefficients in the equation.

PREREQUISITES
  • Understanding of harmonic motion and periodic functions
  • Familiarity with vector notation and operations
  • Knowledge of trigonometric identities, specifically sin²(u) + cos²(u) = 1
  • Basic calculus concepts related to functions and their periods
NEXT STEPS
  • Study the derivation of the period for elliptical motion in physics
  • Learn about vector functions and their applications in motion analysis
  • Explore advanced trigonometric identities and their uses in simplification
  • Investigate the implications of coefficients in periodic functions
USEFUL FOR

Students studying physics, particularly those focusing on mechanics and motion, as well as educators seeking to clarify concepts related to periodic functions and elliptical motion.

Identify
Messages
12
Reaction score
0

Homework Statement



A ball of mass m fastened to a long rubber band is spun around so that the ball follows an elliptical path about the origin given by:

r(t)=bcos(ωt)e(x)+2bsin(ωt)e(y)

b, ω constants
bold type indicates vectors

Find the period of the balls motion.

Homework Equations



r(t)=bcos(ωt)e(x)+2bsin(ωt)e(y)

The Attempt at a Solution


I think the period is, T=2Pi/ω because the motion is harmonic but I'm not sure if this applies for elliptical motion..?

 
Physics news on Phys.org
Identify said:

Homework Statement



A ball of mass m fastened to a long rubber band is spun around so that the ball follows an elliptical path about the origin given by:

r(t)=bcos(ωt)e(x)+2bsin(ωt)e(y)

b, ω constants
bold type indicates vectors

Find the period of the balls motion.

Homework Equations



r(t)=bcos(ωt)e(x)+2bsin(ωt)e(y)

The Attempt at a Solution


I think the period is, T=2Pi/ω because the motion is harmonic but I'm not sure if this applies for elliptical motion..?
Hi Identify, Welcome to Physics Forums.

All you have to do is establish after what time period the function r(t) repeats. What do you know about finding the overall period of a function that is comprised of other functions with their own periods?
 
Thanks gneill.
I think when 2 periodic functions are added their periods are Pi(the lowest common multiple of the two periods). In this case the answer would be T=2Pi/w, since the period of the cos and sin functions are both 2Pi.
 
Identify said:
Thanks gneill.
I think when 2 periodic functions are added their periods are Pi(the lowest common multiple of the two periods). In this case the answer would be T=2Pi/w, since the period of the cos and sin functions are both 2Pi.

Your result is fine.
 
To find the distance from the origin I take,

|r(t)|=((bcos(ωt))^2 + (2bsin(ωt))^2))^1/2

Is there a way I can use the sin^2(u) + cos^2(u) = 1 identity to simplify this any further? Or is this the simplified form? If the identity can be used here I am having trouble with the b and 2b coefficients.
 
Identify said:
To find the distance from the origin I take,

|r(t)|=((bcos(ωt))^2 + (2bsin(ωt))^2))^1/2

Is there a way I can use the sin^2(u) + cos^2(u) = 1 identity to simplify this any further? Or is this the simplified form? If the identity can be used here I am having trouble with the b and 2b coefficients.

Well, you can pull the b out for starters. You could also convert either cos2 or sin2 via your identity to get everything in terms of just sin2 or just cos2.
 

Similar threads

  • · Replies 12 ·
Replies
12
Views
2K
  • · Replies 35 ·
2
Replies
35
Views
4K
  • · Replies 21 ·
Replies
21
Views
3K
  • · Replies 4 ·
Replies
4
Views
2K
Replies
2
Views
1K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 2 ·
Replies
2
Views
17K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 2 ·
Replies
2
Views
5K