Question

A bob of mass 10 kg is attached to a wire 0.3 m long. Its breaking stress is 4.8 x 10⁷ Nm^-2. The area of cross section of the wire is 10^-6 m². What is the maximum angular velocity with which it can be rotated in a horizontal circle?

Answer 1

Since, breaking stress = maximum force applied per unit area.

Given, breaking stress = 4.8 x 10⁷ Nm^-2 and cross-sectional area = 10^-6 m²

Putting the values in above equation we get;

\(4.8\times10^7 = \frac{F_{max}}{10^6\ m^2}\ \therefore\ F_{max} = 48\ N\) 

In the case of horizontal rotation

F_max = mrω²

or 48N = (10 kg)(0.3 m) ω²

ω = 4 rad/s