A model train with a mass of $3 k g$ $3 kg$ is moving along a track at $15 \frac{c m}{s}$ $15 (cm)/s$ . If the curvature of the track changes from a radius of $35 c m$ $35 cm$ to $12 c m$ $12 cm$ , by how much must the centripetal force applied by the tracks change?

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Answer 1 · 2017-07-18T14:20:57

The change in centripetal force is $= 0.37 N$ $=0.37N$

The centripetal force is

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

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

speed, $v = 0.15 m s^{- 1}$ $v=0.15ms^-1$

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

The variation in centripetal force is

$DeltaF=F_2-F_1$

$F_1=mv^2/r_1=3*0.15^2/0.35=0.193N$

$F_2=mv^2/r_2=3*0.15^2/0.12=0.563N$

$DeltaF=0.563-0.193=0.37N$

A model train with a mass of 3 kg3kg3 kg is moving along a track at 15 (cm)/s15cms15 (cm)/s. If the curvature of the track changes from a radius of 35 cm35cm35 cm to 12 cm12cm12 cm, by how much must the centripetal force applied by the tracks change?