A body of mass m = 20 kg lies on a horizontal platform…

A body of mass $m = 20\ \mathrm{kg}$ lies on a horizontal platform rotating with constant angular velocity $\omega = 10\ \mathrm{rad/s}$, at a distance $R = 0.5\ \mathrm{m}$ from the axis. (a) Find the centrifugal force on the body. (b) At what angular velocity does the body start to slide, if the coefficient of friction is $\mu = 0.2$?

Explanation of Method

A body on a rotating platform.

Step 1: (a) Centrifugal force

\[ F_c = m\omega^2 R = 20 \cdot 100 \cdot 0.5. \]

\[F_c = 1000\ \mathrm{N}\]

Step 2: (b) Start of sliding

Static friction holds the body as long as it can supply the needed centripetal force: $m\omega^2 R \le \mu mg$.

\[ \omega \le \sqrt{\frac{\mu g}{R}} = \sqrt{\frac{0.2 \cdot 9.81}{0.5}} = \sqrt{3.92}. \]

\[\omega_{\max} \approx 1.98\ \mathrm{rad/s}\]

$f$ $\text{centrifugal}$ $\omega$ $R = 0.5\ \mathrm{m}$ $\text{holds while } m\omega^2 R \le \mu mg \Rightarrow \omega \le \sqrt{\mu g/R} \approx 1.98\ \mathrm{rad/s}$

Friction from the platform is what keeps the body on its circle. Once the needed $m\omega^2R$ exceeds the largest possible friction $\mu mg$, the body slides outward.

Result

\[\boxed{\,F_c = 1000\ \mathrm{N}\,}\]

\[\boxed{\,\omega_{\max} \approx 1.98\ \mathrm{rad/s}\,}\]

Remark

Note. The original results are correct. They show that the situation in (a) cannot actually happen with $\mu = 0.2$: at 10 rad/s the body would need 1000 N, but friction can give at most 39.2 N, so it would already have slid off. Part (a) only makes sense if the body is held in place some other way.