A model train with a mass of $8 k g$ $8 kg$ 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 $48 c m$ $48 cm$ to $120 c m$ $120 cm$ , by how much must the centripetal force applied by the tracks change?

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Answer 1 · 2017-04-07T10:17:34

The centripetal force changes by $= 0.144 N$ $=0.144N$

The centripetal force is

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

mass, $m = 8 k g$ $m=8kg$

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

radius, $= r$ $=r$

The variation in centripetal force is

$DeltaF=F_2-F_1$

$F_1=8*0.12^2/0.48=0.24N$

$F_2=8*0.12^2/1.20=0.096N$

$DeltaF=0.24-0.096=0.144N$

A model train with a mass of 8 kg8kg8 kg is moving along a track at 12 (cm)/s12cms12 (cm)/s. If the curvature of the track changes from a radius of 48 cm48cm48 cm to 120 cm120cm120 cm, by how much must the centripetal force applied by the tracks change?