Uniform Circular Motion — Question 4

Question 4

Infer the sense of travel from a local tangent

A particle undergoes uniform circular motion about the origin. At one instant its position is $(0,-3)$ m and velocity is $(6,0)$ m/s.

1. Determine clockwise or counterclockwise motion, angular speed and period.

2. Find the acceleration vector at the stated instant.

3. Find position and velocity after one quarter of a revolution, and the elapsed time.

Question 4 - Solution

Infer the sense of travel from a local tangent

(a) Match the tangent to the radius.

$$\omega=v/R=6/3=2\,\mathrm{s^{-1}},\quad\boxed{T=2\pi/\omega=\pi\,\mathrm s}\quad\text{(counterclockwise)}$$ At the bottom of a circle, a rightward tangent corresponds to counterclockwise motion. Radius is 3 m and speed 6 m/s, so $\omega=v/R=2$ s$^{-1}$ and $T=2\pi/\omega=\pi$ s.

(b) Point from the particle toward the center.

$$\vec a=-\omega^2\vec r=-4(0,-3)=(0,12)\,\mathrm{m/s^2}$$ The inward direction is upward. With $a=v^2/R=36/3=12$ m/s$^2$, the vector is $(0,12)$ m/s$^2$. The downward position does not imply downward acceleration.

(c) Advance the radius and tangent together.

$$\Delta t=T/4=\pi/4\,\mathrm s,\quad\vec r_f=(3,0)\,\mathrm m,\quad\vec v_f=(0,6)\,\mathrm{m/s}$$ After a counterclockwise quarter turn, position is $(3,0)$ m and velocity is $(0,6)$ m/s. The elapsed time is $T/4=\pi/4\approx0.7854$ s. Both vector directions have rotated by $90^\circ$ while their magnitudes remain fixed.