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 #1 Hz#, what is the centripetal force acting on the object?

Question

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 #1   Hz#, what is the centripetal force acting on the object?

Layla Ahmed · Accepted Answer

#60π^2N# #"Centripetal force" = (m*v^2)/r# The frequency was given as 1hz Then #"angular velocity" =2πf =(2π" rad")/s# Remember that #"linear velocity" = "angular velocity"*"distance"# #V=romega# Substituting this into the centripetal force formular #F= (m*romega*romega)/r# #F=mrw^2# So using this #F=5 * 3 * 2π * 2π# #F=60 * π^2N#

Charlotte Aaron · Answer

undefined The centripetal force acting on the object can be calculated using the formula:

$$ F = m \cdot r \cdot \omega^2 $$

Where:
- $ F $ is the centripetal force,
- $ m $ is the mass of the object (5 kg),
- $ r $ is the distance from the object to the center of rotation (3 m),
- $ \omega $ is the angular velocity (which is $ 2\pi $ times the frequency, so in this case $ \omega = 2\pi $ radians/s).

Plugging in the values:

$$ F = 5 \times 3 \times (2\pi \times 1)^2 $$

$$ F = 5 \times 3 \times (2\pi)^2 $$

$$ F \approx 94.2 \, \text{N} $$

So, the centripetal force acting on the object is approximately 94.2 N.