A model train, with a mass of #8 kg#, is moving on a circular track with a radius of #1 m#. If the train's rate of revolution changes from #3/8 Hz# to #5/4 Hz#, by how much will the centripetal force applied by the tracks change by?

2 Answers
Mar 10, 2018

We know that angular frequency #omega# and natural frequency #nu# are related as, #omega =2 pinu#

So,if the angular frequency changed from #omega_1# to #omega_2#

we can say, #omega_2 =2pinu_2=2pi(5/4)=7.854 rads^-1# and #omega_1=2pinu_1=2pi(3/8)=2.355 rads^-1#

Now,centripetal force is expressed as #momega^2r#

So,change in centripetal force is #mr(omega_2^2 - omega_1^2)=8*1*(7.854^2-2.355^2)=449.113 N#

Mar 10, 2018

Answer:

The change in centripetal force is #=449.1N#

Explanation:

The centripetal force is

#F=(mv^2)/r=mromega^2N#

The mass of the train, #m=(8)kg#

The radius of the track, #r=(1)m#

The frequencies are

#f_1=(3/8)Hz#

#f_2=(5/4)Hz#

The angular velocity is #omega=2pif#

The variation in centripetal force is

#DeltaF=F_2-F_1#

#F_1=mromega_1^2=mr*(2pif_1)^2=8*1*(2pi*3/8)^2=44.4N#

#F_2=mromega_2^2=mr*(2pif_2)^2=8*1*(2pi*5/4)^2=493.5N#

#DeltaF=F_1-F_2=493.5-44.4=449.1N#