CENTRIPETAL ACCELERATION

Two reasonable ways to show that a_c = v²/r:

1. Consider v at point (x,y) in a circle about the origin, as shown.

The velocity vector v has an angle to the -x axis that is the complement of q , hence

v_x = -vsinq and v_y = vcosq . (equations 1)

But sin and cos of q are y/R and x/R, so

v_x = -vy/R and v_y = vx/R. (equations 2)

Find the rates of change of these and we have a_x and a_y.
Note that v and R are constant, and the rates of change of y and x are v_y and v_x, so

a_x = -vv_y/R and a_y = vv_x/R. (equations 3)

Now substitute equations 2 into equations 3 and get
a_x = -v²x/R² and a_y = -v²y/R²,
and from a = (a_x² + a_y²)^1/2 and x² + y² = R² we find

a = v²/R. Velocity is related to angular velocity w by v = wR if w is in rad/s, so a = w²R is an alternative equation.

Method 2: Any smooth curve has a certain radius of curvature at any given point. (You can find a discussion of radius of curvature in a calculus book. I will take an easier approach.) Since you know about projectile motion (i assume), we will use projectile motion principles. Consider a projectile at the highest point in its motion. The acceleration is straight down, perpendicular to its velocity. Its y component of v is zero at this point.

Now consider its location a small fraction of a second later (call it time t later), small enough time that its path can be
described as a straight line segment of length v_xt. (On any
smooth curve, if you take a small enough portion of it, you
will not be able to tell the difference between that and a
straight line.) Now to be exact, v_y = gt and it is traveling in a direction q below horizontal such that
tanq = gt/v_x.

Since v is perpendicular to r, the velocity vector changes
direction by the same angle as the radius of curvature,
tanq = v_xt/r. If we equate this to the tanq expression in the previous paragraph and multiply both sides by v_x , we get
g = v_x²/r.

By this same reasoning, anytime something is turning with constant speed, the acceleration is toward the center of curvature and is equal to the square of the velocity divided by the radius of curvature.

In general, we could have an acceleration not perpendicular to v. In that case, the component of a perpendicular to v is toward the center and is equal to
v²/r and the component in the v direction is equal to the rate of change of speed. Go accelerate BACK to the other stuff on centripetal acceleration (examples) or to the main mechanics page or to the Newton's 2nd law page.

Other main pages:
fluids, heat, electricity & magnetism
vibrations and waves
quantum
index Comments, questions: fredrick.gram at tri-c.edu (but remove “at” and spaces and insert @)