Uniform Circular Motion — Question 9

Question 9

Revolutions per minute must become a rate per second

A rigid platform turns steadily at 120 revolutions per minute. Point A is 0.30 m from the rotation axis; point B is 0.10 m from it. Both move in circles in the platform plane.

1. Find the frequency, period and angular speed.

2. Find each point’s speed and acceleration magnitude.

3. A calculation substitutes 120 directly for angular speed in $a=\omega^2R$. Explain the two missing conversions.

Question 9 - Solution

Revolutions per minute must become a rate per second

(a) Convert revolutions and time units separately.

$$f=\frac{120}{60}=2\,\mathrm{Hz},\quad T=1/f=.5\,\mathrm s,\quad\boxed{\omega=2\pi f=4\pi\,\mathrm{s^{-1}}}$$ $120\ \mathrm{rev/min}=2\ \mathrm{rev/s}$, so $f=2$ Hz, $T=0.5$ s and $\omega=4\pi$ rad/s. One revolution corresponds to $2\pi$ radians.

(b) Apply the common angular speed at each radius.

$$v_A=1.2\pi,\ v_B=.4\pi\quad\mathrm{m/s};\qquad a_A=4.8\pi^2,\ a_B=1.6\pi^2\quad\mathrm{m/s^2}$$ For A, $v_A=1.2\pi\approx3.770$ m/s and $a_A=4.8\pi^2\approx47.374$ m/s$^2$. For B, $v_B=0.4\pi\approx1.257$ m/s and $a_B=1.6\pi^2\approx15.791$ m/s$^2$. Both ratios A/B are 3.

(c) Audit the unit carried by the rate.

$$\omega=120\,\frac{\mathrm{rev}}{\mathrm{min}}\frac{1\,\mathrm{min}}{60\,\mathrm s}\frac{2\pi\,\mathrm{rad}}{1\,\mathrm{rev}}=4\pi\,\mathrm{rad/s}$$ The stated 120 is in revolutions per minute, not radians per second. Divide by 60 to change minutes to seconds and multiply by $2\pi$ to change revolutions to radians. Squaring an unconverted numerical rate compounds the error. The platform’s points share angular speed, not linear speed.