Mechanics 2: Linear and Rotational Dynamics - 3D Math Primer for Graphics and Game Development - page 607

Imagine a point with mass m that is attached to a rigid disk whose mass
is neglected. The center of the disk is fixed at a pivot, and this constrains
the mass to a circular path. Let r be the vector from the pivot to the mass.
Thus the orbital radius of the mass is r = r . We draw a line on the
disk outward from its center, and we assume that we can a x the mass to
the disk at any distance r along this line. Note that the radius of the disk
itself (being massless) is not relevant, so try to shut off your intuition that a
really big disk would be hard to rotate. The only source of resistance to the
rotation—the only source of inertia—is the point mass. We are neglecting
the inertia of the disk.
Consider what happens when a force f is applied directly at m. Accord-
ing to Newton's second law, the mass wants to accelerate in the direction of
f . However, any portion of f parallel to r will be repelled by a contact force
from the disk and will have no effect on the mass. In contrast, the portion
of f perpendicular to r , tangential to the orbit of the mass, will cause the
point mass to accelerate. Let F denote the magnitude of f , and F ⊥ denote
the amount of f that is perpendicular r . This is depicted in Figure 12.17.

Figure 12.17
A disk is constrained to rotate about o . A
mass m is attached to the disk at a radius r ,
and a force f is applied on the mass.

With a bit of trig, we can compute F ⊥ from φ, which is the angle

between f and r :

F ⊥ /F = sin(π − φ), (π − φ is the interior angle)

F ⊥ /F = sinπ cosφ − cosπ sinφ, (using Equation (1.1))

F ⊥ /F = (0) cosφ − (−1) sinφ,
F ⊥ = F sinφ.

By applying Newton's second law, we can compute the magnitude of the
tangential acceleration as

a ⊥ = F ⊥ /m.