Velocity and Acceleration — Question 3

Question 3

A particle moves on the circle $\vec r(t)=\left\langle R\cos(\omega t),R\sin(\omega t)\right\rangle$, where $R,\omega>0$. Find velocity, acceleration, speed, and show that acceleration points toward the center with magnitude $v^2/R$.

Question 3 – Solution

Strategy Differentiate twice and compare acceleration with the position vector.

Derivatives $$\vec v=R\omega\left\langle -\sin\omega t,\cos\omega t\right\rangle,\qquad \vec a=-R\omega^2\left\langle \cos\omega t,\sin\omega t\right\rangle=-\omega^2\vec r.$$ Thus $\boxed{v=R\omega}$ and acceleration is radially inward.

Magnitude $$\|\vec a\|=R\omega^2=\frac{(R\omega)^2}{R} =\boxed{\frac{v^2}{R}}.$$ Velocity is tangent to the circle and perpendicular to the centripetal acceleration.