A model train with a mass of $3 k g$ $3 kg$ is moving along a track at $16 \frac{c m}{s}$ $16 (cm)/s$ . If the curvature of the track changes from a radius of $15 c m$ $15 cm$ to $32 c m$ $32 cm$ , by how much must the centripetal force applied by the tracks change?

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Answer 1 · 2017-10-30T06:17:28

The change in centripetal force is $= 0.27 N$ $=0.27N$

The centripetal force is

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

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

The speed, $v = (0.16) m s^{- 1}$ $v=(0.16)ms^-1$

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

The variation in centripetal force is

$DeltaF=F_2-F_1$

$F_1=mv^2/r_1=3*0.16^2/0.15=0.512N$

$F_2=mv^2/r_2=3*0.16^2/0.32=0.24N$

$DeltaF=0.512-0.24=0.27N$

A model train with a mass of 3 kg3kg3 kg is moving along a track at 16 (cm)/s16cms16 (cm)/s. If the curvature of the track changes from a radius of 15 cm15cm15 cm to 32 cm32cm32 cm, by how much must the centripetal force applied by the tracks change?