A model train, with a mass of $3 k g$ $3 kg$ , is moving on a circular track with a radius of $8 m$ $8 m$ . If the train's rate of revolution changes from $\frac{1}{4} H z$ $1/4 Hz$ to $\frac{1}{8} H z$ $1/8 Hz$ , by how much will the centripetal force applied by the tracks change by?

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Answer 1 · 2017-11-22T06:43:03

The change in centripetal force is $= 44.4 N$ $=44.4N$

The centripetal force is

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

The mass, $m = (3) k g$ $m=(3)kg$

The radius, $r = 8 m$ $r=8m$

The frequencies are

$f_{1} = \frac{1}{4} = 0.25 H z$ $f_1=1/4=0.25Hz$

$f_{2} = \frac{1}{8} = 0.125 H z$ $f_2=1/8=0.125Hz$

The variation in centripetal force is

$DeltaF=F_2-F_1$

$F_1=mromega_1^2=mr*(2pif_1)^2=3*8*(2pi*0.25)^2=59.2N$

$F_2=mromega_2^2=mr*(2pif_2)^2=3*8*(2pi*0.125)^2=14.8N$

$DeltaF=F_1-F_2=59.2-14.8=44.4N$

A model train, with a mass of 3 kg3kg3 kg, is moving on a circular track with a radius of 8 m8m8 m. If the train's rate of revolution changes from 1/4 Hz14Hz1/4 Hz to 1/8 Hz18Hz1/8 Hz, by how much will the centripetal force applied by the tracks change by?