A body of mass m = 200 g on a rope of length l = 0.5 m…

A body of mass $m = 200\ \mathrm{g}$ on a rope of length $l = 0.5\ \mathrm{m}$ rotates in a vertical plane. The rope breaks when its tension reaches $F_{\max} = 295\ \mathrm{N}$. What is the greatest angular velocity at which the body can rotate without the rope breaking?

Explanation of Method

Greatest angular velocity before a rope breaks.

Step 1: Where the tension is largest

At the lowest point the rope must hold up the weight and also supply the centripetal force: $T - mg = m\omega^2 l \;\Longrightarrow\; T = mg + m\omega^2 l$.

Step 2: Limit

\[ \omega^2 = \frac{F_{\max} - mg}{m\,l} = \frac{295 - 1.962}{0.2 \cdot 0.5} = 2930.4. \]

\[\omega_{\max} \approx 54.1\ \mathrm{rad/s}\]

$T$ $\text{breaks at } T = 295\ \mathrm{N}$ $mg$ $\omega$ $l = 0.5\ \mathrm{m}$ $\text{lowest point:}$ $T - mg = m\omega^2 l$

At the lowest point the rope must hold the weight and also supply the centripetal force, so the tension is greatest there. That is where the rope breaks first.

Result

\[\boxed{\,\omega_{\max} \approx 54.1\ \mathrm{rad/s}\,}\]

Remark

Check. The original result is correct. This is about 8.6 revolutions per second; at this speed the weight (2 N) is tiny compared with the centripetal force (293 N).