A model train, with a mass of $4 k g$ $4 kg$ , is moving on a circular track with a radius of $7 m$ $7 m$ . If the train's rate of revolution changes from $\frac{1}{6} H z$ $1/6 Hz$ to $\frac{1}{3} H z$ $1/3 Hz$ , by how much will the centripetal force applied by the tracks change by?

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Answer 1 · 2018-01-07T07:58:46

The change in centripetal force is $= 92.11 N$ $=92.11N$

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, $m = (4) k g$ $m=(4)kg$

The radius of the track, $r = (7) m$ $r=(7)m$

The frequencies are

$f_{1} = (\frac{1}{6}) H z$ $f_1=(1/6)Hz$

$f_{2} = (\frac{1}{3}) H z$ $f_2=(1/3)Hz$

The variation in centripetal force is

$DeltaF=F_2-F_1$

$F_1=mromega_1^2=mr*(2pif_1)^2=4*7*(2pi*1/6)^2=30.71N$

$F_2=mromega_2^2=mr*(2pif_2)^2=4*7*(2pi*1/3)^2=122.82N$

$DeltaF=F_1-F_2=122.82-30.71=92.11N$

A model train, with a mass of 4 kg4kg4 kg, is moving on a circular track with a radius of 7 m7m7 m. If the train's rate of revolution changes from 1/6 Hz16Hz1/6 Hz to 1/3 Hz13Hz1/3 Hz, by how much will the centripetal force applied by the tracks change by?