Cycloid Particle: Finding Velocity and Acceleration

  • Thread starter Thread starter shyta
  • Start date Start date
  • Tags Tags
    Particle
Click For Summary
SUMMARY

The discussion centers on the motion of a particle following a cycloid path defined by the equations x = Rsin(wt) and y = Rcos(wt) + R, where R and w are constants. Participants explore the calculation of instantaneous velocity and acceleration at the maximum and minimum values of y, specifically at y = 2R and y = 0. The instantaneous velocity at maximum y is confirmed to be 0, while at minimum y, it approaches infinity. The discussion highlights the importance of differentiating the position equations to derive velocity and acceleration components.

PREREQUISITES
  • Understanding of parametric equations in calculus
  • Knowledge of differentiation and its application to motion
  • Familiarity with the concept of cycloid curves
  • Basic vector analysis for velocity and acceleration
NEXT STEPS
  • Study the derivation of parametric equations for cycloid motion
  • Learn about instantaneous velocity and acceleration in vector form
  • Explore the application of calculus in physics, particularly in motion analysis
  • Investigate the properties of cycloidal motion and its applications in mechanics
USEFUL FOR

Students in physics or mathematics, educators teaching calculus and motion, and anyone interested in the dynamics of cycloidal motion.

shyta
Messages
56
Reaction score
0

Homework Statement



A particle moves in a plane according to

x= Rsin(wt) = wRt
y= Rcos(wt) + R

where w and R are constants. This curve, called a cycloid, is the path traced out by a point on the rim of a wheel which rolls without slipping along the x-axis.

Question:
(a) Sketch the path.

(b) Calculate the instantaneous velocity and acceleration when the particle is at its maximum and minimum value of y.


Homework Equations




x= Rsin(wt) = wRt
y= Rcos(wt) + R


The Attempt at a Solution




I drew up the path of the particle, as a cycloid of course. No problem with that.
I am having problems understand part b, with relation to the curve.

I differentiated the given equations wrt to x.

i.e.
dy/dt = -Rw sin (wt)
dx/dt= Rw cos (wt) + wR

then i proceed to find dy/dx = [Rw sin (wt)]/[Rw cos (wt) + wR]


Ymaximum should be 2R (diameter of the rim of the wheel)

I also went ahead to find values of x which corresponds to the maximum and minimum values of y.
Values of X for maximum Y = 0, 2pie, ...
Values of X for minimum Y = pie, 3pie, ...


Instantaneous velocity is tangent of the curve of a position time graph, which is it the same as the graph i drew out?


Then back to the original question, the instantaneous velocity at maximum y is therefore 0? and instantaneous velocity at minimum y is infinity?



I know I have a conceptual error somewhere but I just can't figure it out.
 
Physics news on Phys.org
shyta said:

Homework Statement



A particle moves in a plane according to

x= Rsin(wt) = wRt
y= Rcos(wt) + R


You mean x=Rsin(wt) + wRt, don't you?

shyta said:

Instantaneous velocity is tangent of the curve of a position time graph, which is it the same as the graph i drew out?

Your graph shows y in terms of x. Both the velocity and the acceleration are vectors. vx=dx/dt, vy=dy/dt. The tangent of this graph shows the direction of velocity at the given point. The x and y components of the acceleration are the time-derivatives of vx and vy, respectively.

You have found already that
dy/dt = -Rw sin (wt)
dx/dt= Rw cos (wt) + wR

vx=dx/t and vydy/dt.
At what time instants is y maximum or minimum? Find and plug in the values for t in the equations for vx and vy.

ehild
 

Similar threads

  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 2 ·
Replies
2
Views
1K
Replies
7
Views
2K
  • · Replies 4 ·
Replies
4
Views
3K
Replies
27
Views
4K
Replies
5
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
Replies
25
Views
2K
  • · Replies 40 ·
2
Replies
40
Views
3K
Replies
9
Views
3K