Uniform Circular Motion — Question 1

Question 1

A period determines speed and inward acceleration

A point moves counterclockwise at constant speed around a circle of radius 2 m, completing one revolution every 4 s. For uniform circular motion, $v=2\pi R/T$ and $a=v^2/R$.

1. Find frequency, angular speed, linear speed and acceleration magnitude.

2. At the rightmost point, give velocity and acceleration components in axes centered on the circle.

3. Explain why constant speed does not make acceleration zero.

Question 1 - Solution

A period determines speed and inward acceleration

(a) Convert one revolution into angle and distance.

$$f=1/T=.25\,\mathrm{Hz},\quad\omega=2\pi/T=\pi/2\,\mathrm{s^{-1}},\quad v=\omega R=\pi\,\mathrm{m/s},\quad\boxed{a=\pi^2/2\,\mathrm{m/s^2}}$$ $f=1/T=0.25$ Hz, $\omega=2\pi/T=\pi/2$ s$^{-1}$, $v=\pi$ m/s and $a=\pi^2/2\approx4.935$ m/s$^2$. Radians are dimensionless in the angular-speed units.

(b) Use tangent and inward directions.

$$\vec v=(0,\pi)\,\mathrm{m/s},\quad\vec a=-\omega^2(2,0)=(-\pi^2/2,0)\,\mathrm{m/s^2}$$ At $(2,0)$ m, counterclockwise motion gives $\vec v=(0,\pi)$ m/s. Acceleration is toward the center: $\vec a=(-\pi^2/2,0)$ m/s$^2$. It does not point outward or along the tangent.

(c) Track velocity as a vector.

$$\vec v\cdot\vec a=0\ \Rightarrow\ \frac{d|\vec v|}{dt}=0,\quad\frac{d\vec v}{dt}=\vec a\ne\vec0$$ The magnitude of velocity stays fixed, but its direction changes continuously. Acceleration is the rate of that vector change. Its perpendicular direction preserves speed while turning the velocity toward the next tangent.