A body rotates with frequency nu = 100 Hz along a circle of…

A body rotates with frequency $\nu = 100\ \mathrm{Hz}$ along a circle of radius $r = 1\ \mathrm{m}$. (a) What is its speed? (b) What is its normal acceleration? (c) What angular deceleration stops it after three revolutions? (d) What is the magnitude of its total acceleration just after the braking starts?

Explanation of Method

Speed, normal and total acceleration of a fast-spinning body.

Step 1: (a) Speed

\[ \omega = 2\pi\nu = 200\pi\ \mathrm{rad/s} \approx 628.3\ \mathrm{rad/s}, \]

\[ v = \omega r = 200\pi\ \mathrm{m/s}. \]

\[v \approx 628\ \mathrm{m/s}\]

Step 2: (b) Normal acceleration

\[ a_n = \frac{v^2}{r} = \omega^2 r = (200\pi)^2 \cdot 1 = 40\,000\pi^2\ \mathrm{m/s^2}. \]

\[a_n \approx 3.95 \cdot 10^5\ \mathrm{m/s^2}\]

Step 3: (c) Angular deceleration

Three revolutions are $\theta = 3 \cdot 2\pi = 6\pi\ \mathrm{rad}$. With final angular velocity zero, $0 = \omega_0^2 - 2|\alpha|\theta$: \[|\alpha| = \frac{\omega_0^2}{2\theta} = \frac{(200\pi)^2}{12\pi} = \frac{10\,000\pi}{3}\ \mathrm{rad/s^2}.\]

\[|\alpha| \approx 1.047 \cdot 10^4\ \mathrm{rad/s^2}\]

Step 4: (d) Total acceleration at the start of braking

Now there is also a tangential acceleration, perpendicular to the normal one: $a_t = |\alpha|\, r \approx 1.047 \cdot 10^4\ \mathrm{m/s^2}$.

\[ a = \sqrt{a_n^2 + a_t^2}. \]

\[ a = \sqrt{(3.9478 \cdot 10^5)^2 + (1.0472 \cdot 10^4)^2}\ \mathrm{m/s^2}. \]

\[a \approx 3.949 \cdot 10^5\ \mathrm{m/s^2}\]

$\vec v$ $\vec a_n$ $\vec a_t$ $\vec a$ $O$ $r = 1\ \mathrm{m}$ $a_n = \omega^2 r \approx 3.95 \cdot 10^5\ \mathrm{m/s^2}$ $a_t = |\alpha|\, r \approx 1.05 \cdot 10^4\ \mathrm{m/s^2}$ $a = \sqrt{a_n^2 + a_t^2}$ $\text{tilt: } \arctan\tfrac{a_t}{a_n} \approx 1.5^\circ$

At the moment braking starts. The normal acceleration points to the centre; the tangential acceleration points against the velocity. The tangential part is drawn about 10 times too long: in reality it is only 1/38 of the normal part, so the total acceleration points almost exactly to the centre.

Result

\[\boxed{\,v \approx 628\ \mathrm{m/s}\,}\]

\[\boxed{\,a_n \approx 3.95 \cdot 10^5\ \mathrm{m/s^2}\,}\]

\[\boxed{\,|\alpha| \approx 1.047 \cdot 10^4\ \mathrm{rad/s^2}\,}\]

\[\boxed{\,a \approx 3.949 \cdot 10^5\ \mathrm{m/s^2}\,}\]

Remark

Notes. The tangential part is only $\tfrac{a_t}{a_n} = \tfrac{1}{12\pi} \approx \tfrac{1}{38}$ of the normal part, so braking barely changes the total (by about 0.03 %). The original solution had two small numerical slips: $|\alpha| = 10\,471.98$ (not 10 472.98) rad/s², and $a \approx 394\,923\ \mathrm{m/s^2}$ (not 395 168.89).