# An object with a mass of 5 kg is revolving around a point at a distance of 3 m. If the object is making revolutions at a frequency of 17 Hz, what is the centripetal force acting on the object?

Jan 18, 2016

$1.712 \times {10}^{5} \text{N}$

#### Explanation:

The time period $T$ for 1 revolution $= \frac{1}{f} = \frac{1}{17} s$

$v = \frac{d}{T} = \frac{2 \pi r}{T} = \frac{2 \pi 3}{\frac{1}{17}} = 320.84 \text{m/s}$

The centripetal force is given by:

$F = \frac{m {v}^{2}}{r} = \frac{5 \times {320.84}^{2}}{3} = 1.712 \times {10}^{5} \text{N}$