An object with a mass of #6 kg# is revolving around a point at a distance of #8 m#. If the object is making revolutions at a frequency of #6 Hz#, what is the centripetal force acting on the object?

Question

An object with a mass of #6   kg# is revolving around a point at a distance of #8 m#. If the object is making revolutions at a frequency of #6  Hz#, what is the centripetal force acting on the object?

Sebastian Acosta · Accepted Answer

The force acting on the object is #6912pi^2# Newtons.

We'll start by determining the velocity of the object. Since it is revolving in a circle of radius 8m 6 times per second, we know that: #v = 2pir*6# Plugging in values gives us: #v = 96 pi# m/s Now we can use the standard equation for centripetal acceleration: #a = v^2/r# #a = (96pi)^2/8# #a = 1152pi^2# m/s^2 And to finish the problem we simply use the given mass to determine the force needed to produce this acceleration: #F = ma# #F = 6*1152pi^2# #F = 6912pi^2# Newtons

Carson Alexander · Answer

undefined The centripetal force acting on an object in circular motion can be calculated using the formula:

$$ F_c = m \times r \times (2\pi f)^2 $$

where $ m $ is the mass of the object, $ r $ is the radius of the circular path, and $ f $ is the frequency of revolutions.

Substituting the given values:

$$ F_c = 6 \, \text{kg} \times 8 \, \text{m} \times (2\pi \times 6 \, \text{Hz})^2 $$

$$ F_c = 6 \, \text{kg} \times 8 \, \text{m} \times (2\pi \times 6)^2 \, \text{Hz}^2 $$

$$ F_c = 6 \, \text{kg} \times 8 \, \text{m} \times (36\pi^2) \, \text{Hz}^2 $$

$$ F_c = 1728 \pi^2 \, \text{N} $$

So, the centripetal force acting on the object is $ 1728 \pi^2 \, \text{N} $.