Uniform circular motion

Figure 2.12 shows a particle moving at constant speed υ in a circle of radius r. Suppose that in a short time interval ∆t its position vector rotates through the angle ∆θ and the particle's displacement, ∆ r = r ₂ – r ₁ is vertical. Since v is always perpendicular to r, these two vectors change their directions by the

FIGURE 2.12

|∆v|

same angle in any time interval. In the vector diagram for the equation v ₂ = v ₁ + ∆ v, we know that υ₂ = υ₁ = υ. The direction of ∆ v is horizontal and radially inward–along the bisector of the angle ∆θ drawn within the circle. The triangles OPQ and ABC are both isosceles triangles with the same angles. Thus,

|∆r|

from which we obtain |∆ v | = (υ/r)|∆ r |. Since |∆ r | ≈ υ∆ t, we see that |∆ v |/|∆ t | ≈ υ²/r. From the definition a =lim_{∆ t →0}(|∆ v |/∆ t), we find that the centripetal acceleration is

Centripetal acceleration (2.16)

The subscript r indicates that the acceleration is radial.

The period T is the time it takes to complete one revolution, a distance of 2 πr, so the speed is υ = (2 πr)/ T. Equation 2.16 may be written as

(2.17)

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