A model train with a mass of $5 k g$ $5 kg$ is moving along a track at $12 \frac{c m}{s}$ $12 (cm)/s$ . If the curvature of the track changes from a radius of $32 c m$ $32 cm$ to $24 c m$ $24 cm$ , by how much must the centripetal force applied by the tracks change?

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Answer 1 · 2017-09-27T05:33:51

The change in centripetal force is $= 0.075 N$ $=0.075N$

The centripetal force is

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

The mass, $m = (5) k g$ $m=(5)kg$

The speed, $v = (0.12) m s^{- 1}$ $v=(0.12)ms^-1$

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

The variation in centripetal force is

$DeltaF=F_2-F_1$

$F_1=mv^2/r_1=5*0.12^2/0.32=0.225N$

$F_2=mv^2/r_2=5*0.12^2/0.24=0.3N$

$DeltaF=0.3-0.225=0.075N$

A model train with a mass of 5 kg5kg5 kg is moving along a track at 12 (cm)/s12cms12 (cm)/s. If the curvature of the track changes from a radius of 32 cm32cm32 cm to 24 cm24cm24 cm, by how much must the centripetal force applied by the tracks change?