Derive an expression for the acceleration of a body of mass m moving with a uniform speed v in circular path of radius r.

Explanation

Let A = initial position of the body ,
B = position of the body after a small time interval t.
So, the angular displacement `theta= angleAOB`, is also small, and the displacement x effectively coincides with the arc AB of the circular path. Then,
Now at A, the velocity v is horizontal , at B, the velocity makes the same angle `theta` with the horizontal direction, keeping its magnitude v to be the same, as the body moves with a uniform speed v . Again, as `theta` is small, the change of velocity u, effectively coincides with a circular arc.
So, `theta = (u)/(v)`
Then we get, `(x)/( r) = (u)/(v)`
or, u = `(v)/(r )x`
Hence, the acceleration is,
a = `("change of velocity")/("time") = (u)/(t) = (vx)/(r t)`
= `(v)/(r ) v = (v^(2))/(r )`
also shows that, as `theta` is small, u is normal to the velocity v. As v is tangential at every point on the circular path, the acceleration a is radial. This is centripetal acceleration of a uniform circular motion.