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?

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Answer 1 · 2018-03-10T10:13:14

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 $mr(omega_2^2 - omega_1^2)=8*1*(7.854^2-2.355^2)=449.113 N$

Answer 2 · 2018-03-10T10:16:03

The change in centripetal force is $=449.1N$

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

$DeltaF=F_2-F_1$

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

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

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

A model train, with a mass of 8 kg8 kg, is moving on a circular track with a radius of 1 m1 m. If the train's rate of revolution changes from 3/8 Hz3/8 Hz to 5/4 Hz5/4 Hz, by how much will the centripetal force applied by the tracks change by?