What is the (higher order) time derivative of centripetal acceleration?

Click For Summary
SUMMARY

The time derivative of centripetal acceleration is definitively expressed as ## \vec{r} \omega^3 ##, extending to higher order derivatives as ## \frac {d^n \vec{r}}{dt^n} = \vec{r} \omega^n ##. This conclusion is supported by the analysis presented in Thorton/Marion, specifically on page 390, which discusses rotating bodies in a fixed frame. While dimensional analysis suggests these relationships, it is crucial to perform proper differentiation to validate the results, as shown in the scalar form ## r\omega^2 ## leading to ## r\dot\omega^2 + 2r\omega\dot{\omega} ##.

PREREQUISITES
  • Understanding of centripetal acceleration
  • Familiarity with rotational dynamics
  • Basic knowledge of calculus and differentiation
  • Access to Thorton/Marion textbook for reference
NEXT STEPS
  • Study the implications of higher order time derivatives in rotational motion
  • Review the differentiation of vector functions in dynamics
  • Explore the concepts of angular velocity and its derivatives
  • Analyze examples of rotating bodies from Thorton/Marion for practical applications
USEFUL FOR

This discussion is beneficial for physics students, educators, and professionals in mechanical engineering or applied physics, particularly those focusing on rotational dynamics and centripetal forces.

TheCanadian
Messages
361
Reaction score
13
Just using basic dimensional analysis, it appears the time derivative of centripetal acceleration is ## \vec{r} \omega^3 ##, but this intuitive guess would also extend to higher order time derivatives, no? Implying:

## \frac {d^n \vec{r}}{dt^n} = \vec{r} \omega^n ##

It seems to follow from the general result shown in Thorton/Marion pg 390 (attached) when considering rotating bodies in a fixed frame. I assume ## \vec{Q} ## is any vector, even ones that are the result of a higher order time derivative of an initial vector. The concept of finite infinite-time derivative just seems like an odd concept to me when considering real objects, but I guess the geometry of the situation allows it. But to confirm, is anything posted here incorrect?
 

Attachments

  • Screen Shot 2016-01-16 at 6.20.58 PM.png
    Screen Shot 2016-01-16 at 6.20.58 PM.png
    14.5 KB · Views: 507
Physics news on Phys.org
TheCanadian said:
Just using basic dimensional analysis, it appears the time derivative of centripetal acceleration is r⃗ ω3r→ω3 \vec{r} \omega^3 , but this intuitive guess would also extend to higher order time derivatives, no? Implying:
Dimensional analysis allows that as a possibility, but it does not make it true. If we start with the scalar form ##r\omega^2## we see the time derivative is ##r\dot\omega^2+2r\omega\dot{\omega}##.
 
haruspex said:
Dimensional analysis allows that as a possibility, but it does not make it true. If we start with the scalar form ##r\omega^2## we see the time derivative is ##r\dot\omega^2+2r\omega\dot{\omega}##.

Thank you for pointing out that I forgot to do a basic differentiation. It is much appreciated.
 

Similar threads

  • · Replies 24 ·
Replies
24
Views
4K
  • · Replies 6 ·
Replies
6
Views
4K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 5 ·
Replies
5
Views
10K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 8 ·
Replies
8
Views
2K