In uniform circular motion, the object travels with constant speed v along a fixed circular trajectory with radius r. The speed and radius determine the period of the motion, i.e., the time to complete one full revolution:

V = (circumference / period) = (2πr / T), so, T = (2πr / V)

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We gain insight by considering a numerical example. An object is moving on a circle with radius R = 8.0 cm. The object is at point A at time t = 0, at point B at time t = 2.0 s, and at point C at time t = 4.0 s• The vectors representing the average velocities during the intervals [0, 2 s] and [2 s, 4 s], VBA and VCB , respectively, can be drawn, and the difference vector be found by geometric construction,

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Thus, the acceleration is directed towards the center, so that the net force acting on the object must be directed towards the center as well.

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We find an expression for the magnitude of the acceleration. At time t = 0, the object is at point P1 moving with a velocity v in tangential direction. At time t, the object would be at point P2. Instead, it moves along the circle, and thus is at point P2 . We find an expression for the height using Pythagorean theorem: r2 + (vt)2 = (r + h)2.

 

We assume that the time interval is short, so that vt < r and h < r.

Since (r + h)2 = r2 + 2rh + h2 ≈ r2 ± 2rh, we simplify the RHS: r2 – (vt)2 ≈ r2 ± 2rh so that h = (vt)2/2r = 4 (v2/r)t2. We compare the RHS with x = at2/2, and find the value of the centripetal acceleration:

 ac = (v2/r)

For the numerical example above, we find v2/r ≈ (3 cm/s)2/(6 cm) = 1.5 cm/s2, in agreement with the graphical method.

 

We find the force acting towards the center from Newton’s second law:

Fnet = mac = m (v2 / r)

This is sometimes called “centripetal force;” however, there is no new force involved, and the centriptal force is the net force.

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