Calculate the centripetal force on the end of a \(100{\rm{ m}}\) (radius) wind turbine blade that is rotating at \(0.5{\rm{ rev}}/{\rm{s}}\). Assume the mass is \(4{\rm{ kg}}\).

When a body is moving in a circular path, the net force acts towards the centre which keeps the body moving in a circular path. This force is known as centripetal force.

The angular velocity of the turbine blade is,

\(\omega = 2\pi N\)

Here, \(N\)is the number of revolutions per second.

Substitute \({\rm{0}}{\rm{.5 rev/s}}\) for \(N\),

\(\begin{aligned}{}\omega = 2\pi \times \left( {0.5{\rm{ rev}}/{\rm{s}}} \right)\\ = 3.14{\rm{ rad}}/{\rm{s}}\end{aligned}\)

The centripetal force is,

\(F = m{\omega ^2}r\)

Substitute \({\rm{4 kg}}\) for \(m\), \({\rm{3}}{\rm{.14 rad/s}}\) for \(\omega \), and \({\rm{100 m}}\) for \(r\),

\(\begin{aligned}{}F = \left( {4{\rm{ kg}}} \right) \times {\left( {3.14{\rm{ rad}}/{\rm{s}}} \right)^2} \times \left( {100{\rm{ m}}} \right)\\ = 3.94 \times {10^3}{\rm{ N}}\end{aligned}\)

Hence, the centripetal force is \(3.94 \times {10^3}{\rm{ N}}\).