A model train, with a mass of $2 k g$ $2 kg$ , is moving on a circular track with a radius of $\frac{2}{3} m$ $2/3 m$ . If the train's rate of revolution changes from $\frac{1}{2} H z$ $1/2 Hz$ to $\frac{6}{5} H z$ $6/5 Hz$ , by how much will the centripetal force applied by the tracks change by?

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Answer 1 · 2018-01-01T06:49:51

The change in centripetal force is $= 62.6 N$ $=62.6N$

The centripetal force is

$F = \frac{m v^{2}}{r} = m r ω^{2} N$ $F=(mv^2)/r=mromega^2N$

The mass of the train, $m = (2) k g$ $m=(2)kg$

The radius of the track, $r = (\frac{2}{3}) m$ $r=(2/3)m$

The frequencies are

$f_{1} = (\frac{1}{2}) H z$ $f_1=(1/2)Hz$

$f_{2} = (\frac{6}{5}) H z$ $f_2=(6/5)Hz$

The variation in centripetal force is

$DeltaF=F_2-F_1$

$F_1=mromega_1^2=mr*(2pif_1)^2=2*2/3*(2pi*1/2)^2=13.2N$

$F_2=mromega_2^2=mr*(2pif_2)^2=2*2/3*(2pi*6/5)^2=75.8N$

$DeltaF=F_1-F_2=75.8-13.2=62.6N$

A model train, with a mass of 2 kg2kg2 kg, is moving on a circular track with a radius of 2/3 m23m2/3 m. If the train's rate of revolution changes from 1/2 Hz12Hz1/2 Hz to 6/5 Hz65Hz6/5 Hz, by how much will the centripetal force applied by the tracks change by?