Mathematics of Physics and Engineering

(^36) Curves in Space
The velocity v is a sum of the radial velocity component fr and the angular
velocity component rujd. We call f and r6 the radial and angular speeds,
respectively.
The acceleration a in the frame O is obtained by differentiating (1.3.24)
with respect to t according to the rules (1.3.3), (1.3.4):
a = dv/dt = rr + f (dr/d0)6 + (rd + r$)9 + r6 (dO/d9)9,
or
a = (r-r6^2 )r + (r6 + 2f6)6. (1.3.25)
The acceleration a is a sum of the radial component ar and the angular
component ag, where
ar = (f - ruj^2 ) r and ae = {r§ + 2fuj) 6. (1.3.26)
EXERCISE 1.3.14.B Verify that decomposition (1.3.26) of the acceleration
is a particular case of (1.3.19).
Now assume that the trajectory of the point mass is a circle with center
at O and radius R. Then r(t) = R for all t and f(t) = r(t) = 0. Let
9\t) =u(t). By (1.3.26),
{ (1.3.27)
ar = —RUJ^2 f (centripetal acceleration)
ag = Rwd (angular acceleration).
Also, by (1.3.24),
v = Rwd. (1.3.28)
EXERCISE 1.3.15.^5 Verify that formula (1.3.27) is a particular case of the
decomposition (1.3.21) of the acceleration, as derived on page 34-
If we further assume that the angular speed is constant, that is, u>(t) =
u>o for all t, then w = 0, and, by (1.3.27),
ar = -Ru%r, a 0 = 0. (1.3.29)
EXERCISE 1.3.16.B Verify that if the acceleration of a point mass in polar
coordinates is given by (1.3.29), then the point moves around the circle of
radius R with constant angular speed WQ. Hint: Combine (1.3.29) and (1.3.26)
to get differential equations for r and 6. Solve the equations with initial conditions
r(0) = R, r(0) = 0, 0(0) = 0, 0(0) = w 0 to get r{i) = R, 0(i) = w 0 t.