Centripetal Acceleration derivation help

Click For Summary
SUMMARY

The discussion focuses on the derivation of centripetal acceleration, specifically addressing the relationship between velocity and acceleration in circular motion. The key equations presented include \( w = \frac{2\pi}{T} \), \( v = rw = \frac{2\pi r}{T} \), and \( a = \frac{v^2}{R} \). A critical error identified is the incorrect differentiation of velocity with respect to time, as the magnitude of velocity remains constant while its direction changes continuously. The correct approach involves using vector calculus to derive acceleration, confirming that \( a = \frac{v^2}{r} \).

PREREQUISITES
  • Understanding of circular motion and centripetal acceleration
  • Familiarity with angular velocity and its relation to linear velocity
  • Basic knowledge of vector calculus
  • Proficiency in differentiating functions with respect to time
NEXT STEPS
  • Study the derivation of centripetal acceleration using vector calculus
  • Learn about angular velocity and its applications in circular motion
  • Explore the relationship between linear and angular quantities in physics
  • Investigate the implications of constant velocity in circular motion on acceleration
USEFUL FOR

Students of physics, educators teaching mechanics, and anyone interested in understanding the principles of circular motion and centripetal acceleration.

qazxsw11111
Messages
95
Reaction score
0
w=2π /T
v=rw=2πr/T
a=rw2=r(4π2r2/T)

This I know, but why can't I just differentiate v with respect to t to get a? (But the answer is wrong) Can anyone tell me why?

Thanks.
 
Physics news on Phys.org
Also, a_{c}=\frac{v^{2}}{R} in case you didn't have this equation at your disposal.

Are you trying to differentiate v=rw=2πr/T with respect to time?
 
w=2π /T
v=rw=2πr/T
a=rw²=r(4π²r²/T)
Error in the third line - should be a = a=rw²=r(2π /T)² = 4π²r/T²
Differentiating v = 2πr/T does no good because this formula is for the (constant) magnitude of velocity, so the dv/dt = 0. In fact the direction of the velocity is continuously changing so a vector derivative must be done to get the acceleration that way. It would be something like this:
v = 2πr/T[cos(ωt),sin(ωt)]
dv/dt = 2πr/T[-ω*sin(ωt), ω*cos(ωt)]
= 2πr/T*ω[-sin(ωt), cos(ωt)] and since ω = v/r and v = 2πr/T this is
= v²/r[-sin(ωt), cos(ωt)]
showing that the magnitude of acceleration is good old v²/r and that its direction rotates with time.
 

Similar threads

Replies
3
Views
3K
Replies
8
Views
1K
  • · Replies 28 ·
Replies
28
Views
3K
Replies
55
Views
3K
Replies
2
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 8 ·
Replies
8
Views
3K
  • · Replies 5 ·
Replies
5
Views
3K