A model train with a mass of $4 k g$ $4 kg$ is moving along a track at $18 \frac{c m}{s}$ $18 (cm)/s$ . If the curvature of the track changes from a radius of $25 c m$ $25 cm$ to $15 c m$ $15 cm$ , by how much must the centripetal force applied by the tracks change?

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Answer 1 · 2017-05-21T17:08:05

The change in centripetal force is $= 0.34 N$ $=0.34N$

The centripetal force is

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

mass, $m = 4 k g$ $m=4kg$

speed, $v = 0.18 m s^{- 1}$ $v=0.18ms^-1$

radius, $= (r) m$ $=(r) m$

The variation in centripetal force is

$DeltaF=F_2-F_1$

$F_1=mv^2/r_1=4*0.18^2/0.25=0.52N$

$F_2=mv^2/r_2=4*0.18^2/0.15=0.86N$

$DeltaF=0.86-0.52=0.34N$

A model train with a mass of 4 kg4kg4 kg is moving along a track at 18 (cm)/s18cms18 (cm)/s. If the curvature of the track changes from a radius of 25 cm25cm25 cm to 15 cm15cm15 cm, by how much must the centripetal force applied by the tracks change?