Uniform Circular Motion — Question 7

Question 7

Recover a circle from acceleration and period

A particle in uniform circular motion has acceleration magnitude 18 m/s$^2$ and period $2\pi/3$ s. Its circle is centered at $(1,-2)$ m, and its motion is counterclockwise.

1. Find angular speed, radius and linear speed.

2. At the circle’s rightmost point, find position, velocity and acceleration vectors.

3. Explain why $\vec a=-\omega^2\vec r$ would be wrong for the stated origin, and give the corrected vector relation.

Question 7 - Solution

Recover a circle from acceleration and period

(a) Use the two independent rate relations.

$$\omega=\frac{2\pi}{2\pi/3}=3\,\mathrm{s^{-1}},\quad\boxed{R=a/\omega^2=2\,\mathrm m,\quad v=\omega R=6\,\mathrm{m/s}}$$ $\omega=2\pi/T=3$ s$^{-1}$. Since $a=\omega^2R$, $R=18/9=2$ m. Then speed is $v=\omega R=6$ m/s. The period and acceleration together fix the radius.

(b) Add the center offset only to position.

$$\vec r=\vec r_C+(R,0)=(3,-2)\,\mathrm m,\quad\vec v=(0,6)\,\mathrm{m/s},\quad\vec a=(-18,0)\,\mathrm{m/s^2}$$ The rightmost position is $(3,-2)$ m. Velocity there is $(0,6)$ m/s and acceleration is $(-18,0)$ m/s$^2$. Tangent and inward directions are defined relative to the circle’s center, not the coordinate origin.

(c) Use the radius vector from the actual center.

$$\vec a=-\omega^2(\vec r-\vec r_C)=-9(2,0)=(-18,0)\,\mathrm{m/s^2}\ne-9(3,-2)$$ The correct equation is $\vec a=-\omega^2(\vec r-\vec r_C)$, with $\vec r_C=(1,-2)$ m. Applying $-\omega^2\vec r$ would point toward the origin rather than toward C and give the wrong magnitude. A fixed origin shift changes positions but not acceleration.