There's a nice vector way to do this for a particle moving in a circle of **constant** radius with position #vec r(t)# at **uniform** speed #|vec v(t)|#

we can say that

#v^2 = vec v * vec v = const#

so differentiating wrt time using product rule for the vectors

# vec dot v * vec v + vec v * vec dot v = 0 = 2 vec v vec dot v qquad star#

and vecause

#r^2 = vec r * vec r = const#

we can do the same

# vec dot r * vec r + vec r * vec dot r = 0 = 2 vec r vec dot r qquad triangle#

#triangle# amounts to # vec r * vec v = 0#

#star# amounts to # vec v * vec a = 0#

this means, ignoring obvious trivial conclusions, that #vec r# and #vec v# are perpendicular, which we already know from the circular motion, but also that #vec v# and #vec a# are perpendicular

which means that #vec r# and #vec a# are pointing in the same direction or opposite directions which means we already know that the particle is accelerating radially inward out outward. **so there must be a radial force acting to cause the acceleration** - Newton"s Second Law.

if we differentiate #triangle# again using the product rule, we get

# vec dot r * vec v + vec r * vec dot v= 0#

or

# vec v * vec v + vec r * vec a= 0#

or

# vec r * vec a= - v^2 qquad = |vec r| |vec a| cos psicirc#

so we now conclude from the definition of the dot product that #psi = - pi#, acceleration is inwardly radial, and in scalar terms #a = v^2 /r#

So from Newton's Second Law

#F = (m v^2 )/r#

or

#vec F = - (m v^2 )/r \ hat r#