An object with a mass of #2 kg# is revolving around a point at a distance of #4 m#. If the object is making revolutions at a frequency of #5 Hz#, what is the centripetal force acting on the object?

Question

An object with a mass of #2    kg# is revolving around a point at a distance of #4  m#. If the object is making revolutions at a frequency of #5   Hz#, what is the centripetal force acting on the object?

Isabella Alvarez · Accepted Answer

#F=1600*pi^2N# #"you can use the Newton's second law"# #F=m*a# #"change linear acceleration to rotary acceleration"# #a=v^2/r ; v=omega*r ; v^2=omega^2*r^2# #a=omega^2*r# #F=m*omega^2*r# #F=m*(2*pi*f)^2*r# #F=4*4*pi^2*25*4# #F=4^3*5^2*pi^2# #F=64*25*pi^2# #F=1600*pi^2N#

Julia Aguilar · Answer

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

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

where:
- $ F $ is the centripetal force,
- $ m $ is the mass of the object (2 kg),
- $ r $ is the distance from the center of rotation (4 m),
- $ f $ is the frequency of revolution (5 Hz), and
- $ \pi $ is the mathematical constant pi (approximately 3.14159).

Substituting the given values into the formula:

$ F = 2 \, \text{kg} \times 4 \, \text{m} \times (2\pi \times 5 \, \text{Hz})^2 $

$ F = 2 \times 4 \times (2\pi \times 5)^2 $

$ F = 2 \times 4 \times (10\pi)^2 $

$ F = 2 \times 4 \times 100\pi^2 $

$ F = 800\pi^2 $ N

Calculating $ 800\pi^2 $ yields approximately 7853.98 N. So, the centripetal force acting on the object is approximately 7854 N.