Acceleration in circular motion

  • Thread starter oneplusone
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  • #1
oneplusone
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Hello,
Regarding acceleration in circular motion, my textbook says the total acceleration
of an object traveling in a circular path, can be computed by:

[itex] a = \sqrt{a^2_c + a^2_t} [/itex]

and can be proved by pythag. thm.
Can someone help me understand this intuitively?
Thanks.
 

Answers and Replies

  • #2
robphy
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At each instant, the Acceleration Vector is nonzero when the Velocity Vector changes.

The component of the acceleration that is perpendicular to the Velocity Vector, [itex]a_c[/itex], is associated with changing the direction of the Velocity Vector (turning the velocity vector without changing its magnitude).

The component of the acceleration that is parallel to the Velocity Vector, [itex]a_t[/itex], is associated with changing the magnitude of the Velocity Vector (speeding up or slowing down, without changing its direction).

The total acceleration vector is the vector sum of these two components.
Since they are perpendicular to each other, you use the Pythagorean Theorem to compute the magnitude of the Acceleration Vector.
 

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