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

1 Answer
Aug 19, 2017

Answer:

#F_c~~5053N# radially inward.

Explanation:

The centripetal force is given in accordance with Newton's second law as:

#F_c=ma_c#

where #m# is the mass of the object and #a_c# is the centripetal acceleration experienced by the object

The centripetal acceleration is given by:

#a_c=v^2/r#

which is equivalent to #romega^2#. Therefore, we can write:

#F_c=mromega^2#

where #r# is the radius and #omega# is the angular velocity of the object

The angular velocity can also be expressed as:

#omega=2pif#

where #f# is the frequency of the revolution

And so our final expression becomes:

#color(blue)(F_c=mr(2pif)^2#

We are given:

  • #|->"m"=2"kg"#

  • #|->"r"=4"m"#

  • #|->f=4"s"^-1#

Substituting these values into the equation we derived above:

#F_c=(2"kg")(4"m")(2pi(4"s"^-1))^2#

#=5053.237"N"#

#~~5053N# radially inward

This may also be expressed in scientific notation as #5xx10^3"N"# where significant figures are concerned.