A model train, with a mass of $8 k g$ $8 kg$ , is moving on a circular track with a radius of $1 m$ $1 m$ . If the train's rate of revolution changes from $\frac{5}{8} H z$ $5/8 Hz$ to $\frac{5}{4} H z$ $5/4 Hz$ , by how much will the centripetal force applied by the tracks change by?

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Answer 1 · 2017-03-04T20:50:42

$123.37 N$ $123.37N$

The centripetal force of an object is given by

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

You can also find $v$ $v$ by

$v = 2 π r f$ $v = 2pirf$

where $f$ $f$ is the frequency of revolution.

Substitute this into the above equation,

$F = \frac{m {(2 π r f)}^{2}}{r} = \frac{4 m π^{2} r^{2} f^{2}}{r}$ $F = (m(2pirf)^2)/r = (4mpi^2r^2f^2)/r$

$= 4 m r π^{2} f^{2}$ $=4mrpi^2f^2$

or, since we're looking at the change in centripetal force from the change in frequency, then

$DeltaF = 4mrpi^2(Deltaf)^2$

where $Deltaf$ is $5/4Hz - 5/8Hz = 5/8Hz$

Therefore, using the values that we know,

$DeltaF = 4xx8xx1xxpi^2xx(5/8)^2$

$= 123.37N$

A model train, with a mass of 8 kg8kg8 kg, is moving on a circular track with a radius of 1 m1m1 m. If the train's rate of revolution changes from 5/8 Hz58Hz5/8 Hz to 5/4 Hz54Hz5/4 Hz, by how much will the centripetal force applied by the tracks change by?