6.3 Circular Motion: Tangential and Radial Acceleration When the motion of an object is described in polar coordinates, the acceleration has two components, the tangential component aθ , and the radial component, ar . We can write the acceleration vector as ! ˆ . a = ar r̂(t) + aθ θ(t) (6.3.1) ˆ Keep in mind that as the object moves in a circle, the unit vectors r̂(t) and θ(t) change direction and hence are not constant in time. We will begin by calculating the tangential component of the acceleration for circular motion. Suppose that the tangential velocity vθ = r dθ / dt is changing in magnitude due to the presence of some tangential force; we shall now consider that dθ / dt is changing in time, (the magnitude of the velocity is changing in time). Recall that in polar coordinates the velocity vector Eq. (6.2.8) can be written as dθ ! v (t) = r θ̂(t) . dt (6.3.2) We now use the product rule to determine the acceleration. ! dθ (t) dθ̂(t) dv (t) d 2θ (t) ! a(t) = =r θ̂(t) + r . 2 dt dt dt dt (6.3.3) Recall from Eq. (6.2.3) that θ̂(t) = − sin θ (t) î + cosθ (t) ĵ . So we can rewrite Eq. (6.3.3) as d 2θ (t) dθ (t) d ! a(t) = r θ̂(t) + r (− sin θ (t) î + cosθ (t) ĵ) . 2 dt dt dt (6.3.4) We again use the chain rule (Eqs. (6.2.5) and (6.2.6)) and find that d 2θ (t) ˆ dθ (t) ⎛ dθ (t) dθ (t) ! θ(t ) + r a(t ) = r î − sin θ (t) ⎜⎝ − cosθ (t) 2 dt dt dt dt ⎞ ĵ⎟ . ⎠ (6.3.5) 1 Recall that ω ≡ dθ / dt , and from Eq. (6.2.2), r̂(t) = cosθ (t) î + sin θ (t) ĵ , therefore the acceleration becomes 2 d 2θ (t) ! ⎛ dθ (t) ⎞ θ̂(t) − r ⎜ r̂(t) . a(t) = r (6.3.6) ⎝ dt ⎟⎠ dt 2 The tangential component of the acceleration is then aθ = r d 2θ (t) . dt 2 (6.3.7) The radial component of the acceleration is given by 2 ⎛ dθ (t)⎞ ar = −r ⎜ = −r ω 2 < 0 . ⎟ ⎝ dt ⎠ (6.3.8)  Because ar < 0 , that radial vector component a r (t) = −r ω 2 r̂(t) is always directed towards the center of the circular orbit. Example 6.1 Circular Motion Kinematics A particle is moving in a circle of radius R . At t = 0 , it is located on the x -axis. The angle the particle makes with the positive x -axis is given by θ (t ) = At 3 − Bt , where A and B are positive constants. Determine (a) the velocity vector, and (b) the acceleration vector. Express your answer in polar coordinates. At what time is the centripetal acceleration zero? Solution: The derivatives of the angle function θ (t) = At 3 − Bt are dθ / dt = 3At 2 − B and d 2θ / dt 2 = 6 At . Therefore the velocity vector is given by dθ (t) ! θ̂(t) = R(3At 2 − Bt)θ̂(t) . v (t) = R dt 2 The acceleration is given by d 2θ (t)  ⎛ dθ (t) ⎞ θ̂(t) − R ⎜ a(t) = R r̂(t) 2 ⎝ dt ⎟⎠ dt . 2 ( = R(6At) θ̂(t) − R 3At 2 − B ) 2 r̂(t) The centripetal acceleration is zero at time t = t1 when 3At12 − B = 0 ⇒ t1 = B / 3A . 3