An object with a mass of #5 kg# is revolving around a point at a distance of #2 m#. If the object is making revolutions at a frequency of #4 Hz#, what is the centripetal force acting on the object?
1 Answer
Answer:
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
#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"=5"kg"# 
#"r"=2"m"# 
#f=4"s"^1#
Substituting these values into the equation we derived above:
#F_c=(5"kg")(2"m")(2pi(4"s"^1))^2#
#=6316.547"N"#
#~~6317N# radially inward