How to find tangential and normal acceleration?

Click For Summary
SUMMARY

The discussion focuses on calculating tangential and normal acceleration from given acceleration equations, specifically $$a_x=c*cos[d*t]$$ and $$a_y=c*sin[d*t]$$. Participants reference equations for tangential acceleration $$a_t=(r\ddot{\theta}+2\dot{r}\dot{\theta})$$ and normal acceleration $$a_n=(\ddot{r}-r\dot{\theta}^2$$, while also considering an alternative form $$a_t=\frac{dv}{dt}|v|$$. The confusion arises around the application of these equations without a defined angle theta, leading to a need for clarification on how to derive tangential acceleration accurately.

PREREQUISITES
  • Understanding of polar coordinates and their derivatives
  • Familiarity with the concepts of tangential and normal acceleration
  • Knowledge of calculus, specifically differentiation and integration
  • Basic grasp of kinematics and motion equations
NEXT STEPS
  • Study the derivation of tangential and normal acceleration in polar coordinates
  • Learn how to apply the chain rule in the context of acceleration
  • Explore the relationship between velocity and acceleration in circular motion
  • Investigate the use of parametric equations in motion analysis
USEFUL FOR

Students and educators in physics or engineering, particularly those focusing on dynamics and motion analysis, will benefit from this discussion.

reslion
Messages
2
Reaction score
0

Homework Statement


The problem asks for the tangential and normal acceleration of the acceleration. We were given that:
$$a_x=c*cos[d*t]$$ and $$a_y=c*sin[d*t]$$ where c and d are constants.

Homework Equations


The book gives us
$$a_t=(r\ddot{\theta}+2\dot{r}\dot{\theta})$$, (1)
$$a_n=(\ddot{r}-r\dot{\theta}^2$$ (2)
and
$$a=\sqrt{(a_t)^2+(a_n)^2}$$ (3)
but I found online that
$$a_t=\frac{dv}{dt}|v|$$ (4).
Finally, I know that $$a=\sqrt{(a_x)^2+(a_y)^2}$$ (5).

The Attempt at a Solution


My attempt was to use equation (4) as we don't have a theta to find $$a_{tx}$$ and $$a_{ty}$$. But wouldn't that just be the given accelerations as we would integrate the acceleration just to take the derivative again?
After that I would just solve (3) for $$a_n$$ of x and y and then plug both it in the (5) to find the magnitude of both $$a_t$$ and $$a_n$$. But I'm confused on how to find the tangential with either (1) or (4)?
 
Physics news on Phys.org
I don't understand your equation 3. A correct version of that would be useful.
Remember that the tangential acceleration is, by definition, the component of the acceleration in the direction of the velocity.
 

Similar threads

  • · Replies 13 ·
Replies
13
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
6
Views
1K
Replies
5
Views
3K
Replies
4
Views
2K
  • · Replies 20 ·
Replies
20
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
Replies
8
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K