A model train with a mass of $3 k g$ $3 kg$ is moving along a track at $8 \frac{c m}{s}$ $8 (cm)/s$ . If the curvature of the track changes from a radius of $15 c m$ $15 cm$ to $18 c m$ $18 cm$ , by how much must the centripetal force applied by the tracks change?

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Answer 1 · 2017-04-26T16:51:08

The centripetal force changes by $= 0.021 N$ $=0.021N$

The centripetal force is

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

mass, $m = 3 k g$ $m=3kg$

speed, $v = 0.08 m s^{- 1}$ $v=0.08ms^-1$

radius, $= r$ $=r$

The variation in centripetal force is

$DeltaF=F_2-F_1$

$F_1=mv^2/r_1=3*0.08^2/0.15=0.128N$

$F_2=mv^2/r_2=3*0.08^2/0.18=0.107N$

$DeltaF=0.128-0.107=0.021N$

A model train with a mass of 3 kg3kg3 kg is moving along a track at 8 (cm)/s8cms8 (cm)/s. If the curvature of the track changes from a radius of 15 cm15cm15 cm to 18 cm18cm18 cm, by how much must the centripetal force applied by the tracks change?