Two-Dimensional Motion of a Particle: Velocity & Acceleration

  • Thread starter Thread starter chart2006
  • Start date Start date
  • Tags Tags
    Motion Particle
Click For Summary
SUMMARY

The discussion focuses on solving the two-dimensional motion of a particle defined by the equations r = 1/(sinθ - cosθ) and tanθ = 1 + 1/t². Participants aim to determine the magnitudes of velocity and acceleration, as well as the radius of curvature of the path. A key insight is that instead of solving for θ directly, one should derive expressions for sinθ and cosθ in terms of time t, leveraging the properties of right triangles. This approach simplifies the differentiation process needed to find the required motion parameters.

PREREQUISITES
  • Understanding of polar coordinates and their relationships
  • Knowledge of differentiation in calculus
  • Familiarity with trigonometric functions and their properties
  • Basic concepts of kinematics in two-dimensional motion
NEXT STEPS
  • Study the derivation of velocity and acceleration in polar coordinates
  • Learn how to calculate the radius of curvature for parametric equations
  • Explore the relationship between trigonometric identities and motion equations
  • Practice solving similar problems involving two-dimensional motion
USEFUL FOR

Students studying physics or engineering, particularly those focusing on kinematics and motion analysis in two dimensions.

chart2006
Messages
11
Reaction score
0

Homework Statement



The two-dimensional motion of a particle is defined by the relationship [tex]r = \frac {1}{sin\theta - cos\theta}[/tex] and [tex]tan\theta = 1 + \frac {1}{t^2}[/tex], where [tex]r[/tex] and [tex]\theta[/tex] are expressed in meters and radians, respectively, and [tex]t[/tex] is expressed in seconds. Determine (a) the magnitudes of velocity and acceleration at any instant, (b) the radius of curvature of the path.


Homework Equations



[tex]r = \frac {1}{sin\theta - cos\theta}[/tex]


[tex]tan\theta = 1 + \frac {1}{t^2}[/tex]



The Attempt at a Solution



I've made a few attempts but they seem way more complicated than the problem should be I think. I'm assuming I need to solve [tex]tan\theta[/tex] for [tex]\theta[/tex]. Once I've done that I figure I'd need to differentiate both [tex]r[/tex] and [tex]\theta[/tex] to find [tex]\dot{r}, \ddot{r}, \dot{\theta}, \ddot{\theta}[/tex].

I don't know if I'm on the correct route but any help would be appreciated. thanks!
 
Physics news on Phys.org
chart2006 said:
I've made a few attempts but they seem way more complicated than the problem should be I think. I'm assuming I need to solve [tex]tan\theta[/tex] for [tex]\theta[/tex].

Tangent is not a one-to-one function, so that's a bad idea.

Instead, draw a picture!:smile: I think you can find expressions for [itex]\sin\theta[/itex] and [itex]\cos\theta[/itex] in terms of [itex]t[/itex] without actually solving for [itex]\theta[/itex] first...think 'right triangle':wink:
 

Similar threads

Replies
6
Views
2K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 9 ·
Replies
9
Views
2K
Replies
2
Views
3K
  • · Replies 11 ·
Replies
11
Views
3K
Replies
10
Views
2K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K