What is the computation of tangential acceleration in circular motion?

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SUMMARY

The computation of tangential acceleration in circular motion involves understanding both tangential and centripetal acceleration. In this discussion, the automobile accelerates at a rate of 0.600 m/s² while traveling along a circular path with a radius of 20.0 m. The tangential acceleration is confirmed to be 0.600 m/s², as it represents the rate of change of speed. Additionally, the centripetal acceleration, which is necessary to maintain circular motion, is calculated using the instantaneous speed of 3.00 m/s.

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  • Understanding of circular motion principles
  • Knowledge of tangential and centripetal acceleration
  • Familiarity with basic kinematic equations
  • Ability to perform vector calculations
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lostinphys
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I'm sorry guys... I'm stuck again...
An automobile whose speed is increasing at a rate of 0.600 m/s2 travels along a circular road of radius 20.0 m. When the instantaneous speed of the automobile is 3.00 m/s, find the tangential acceleration component.
Magnitude.

... i know that the tangential acceleration = radius (angular acceleration) , but i am having trouble understanding the actual computation of tangential acceleration. i picked up a couple of books, and I'm still pretty lost.
 
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lostinphys said:
I'm sorry guys... I'm stuck again...
An automobile whose speed is increasing at a rate of 0.600 m/s2 travels along a circular road of radius 20.0 m. When the instantaneous speed of the automobile is 3.00 m/s, find the tangential acceleration component.
Magnitude.

... i know that the tangential acceleration = radius (angular acceleration) , but i am having trouble understanding the actual computation of tangential acceleration. i picked up a couple of books, and I'm still pretty lost.


hi there...

im new to this site.. sorry i can't help you w/ that problem... maybe you can help me with a simple question i have?
 
Centripetal acceleration does not affect speed (it changes the direction of velocity).
I'd say the tangental acceleration is 0,6 m/s^2.


If 0,6 m/s^2 was, however, meant as the total acceleration:
If the instantaneous velocity at the given instant t is 3,00 m/s, what is the centripetal acceleration needed to keep the object on its circular orbit of 20m radius?
Tangental and centripetal acceleration are perpendicular. The given total acceleration is the hypotenuse.
 

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