Perpendicular Velocity and Displacement in Particle Motion

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SUMMARY

The discussion centers on the mathematical proof that a particle's path lies on a sphere when its displacement and velocity vectors are always perpendicular. The participant demonstrates this by expressing displacement as r(t) = xi + yj + zk and showing that r(t) · v(t) = 0 leads to the conclusion that the particle's motion is constrained to a spherical surface. Additionally, it is established that if the particle moves with constant speed, the velocity and acceleration vectors remain perpendicular, as the acceleration can be expressed in terms of curvature and normal vectors, confirming their orthogonality.

PREREQUISITES
  • Understanding of vector calculus, specifically displacement and velocity vectors.
  • Knowledge of spherical geometry and equations of spheres.
  • Familiarity with concepts of curvature and normal vectors in motion.
  • Intermediate calculus skills, particularly in differentiating vector functions.
NEXT STEPS
  • Study the derivation of the equation of a sphere in three-dimensional space.
  • Learn about the properties of curvature and its implications in particle motion.
  • Explore the relationship between velocity, acceleration, and curvature in vector calculus.
  • Investigate the implications of constant speed on the motion of particles in physics.
USEFUL FOR

Students and educators in mathematics and physics, particularly those focusing on calculus, vector analysis, and kinematics in particle motion.

tifa8
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Hi ! I have a intemediate calculus problem. I am seeing right now all about curves and motion curves.

Homework Statement



Show that 1. the path of a particle lies on a sphere if its displacement and velocity are always perpendicular
2.show that if the particle moves with constant speed then the velocity and acceleration are penperdicular.

Homework Equations





The Attempt at a Solution



1.I think that it is obvious that it lies on a sphere. however I really don't know how to demonstrate that...

if i consider the displacement as r(t)=xi+yj+zk
then r(t).v(t)=0
=>r(t).r'(t)=0
=>xx'+yy'+zz'=0

but it is totally different from a sphere equation which is x^2+y^2+z^2=R^2

2. I don't know if my reasoning is true
since v is constant then v' is equal to zero, thus a is equal to zero

so a.v=0.v=0 so a and v are perpendicular. However, a is a zero vector which I think is quite strange...

thank you for your help
 
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I think I found the answer for question 2.

vector v=Su where u is a unit tangent vector
since the speed S is constant, then S'=0

Or vector a= S'u+K(s)^2N where N is a unit normal vector and K is the curvature
=> a=K(s)^2N

so a.v=K(s)^2N.Su=K(s)^3 N.u=0 since N.u=0 (perpendicular to each other)
 

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