A model train with a mass of $4 k g$ $4 kg$ is moving along a track at $6 \frac{c m}{s}$ $6 (cm)/s$ . If the curvature of the track changes from a radius of $56 c m$ $56 cm$ to $48 c m$ $48 cm$ , by how much must the centripetal force applied by the tracks change?

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Answer 1 · 2017-03-10T13:25:16

The change in centripetal force is $= 0.0043 N$ $=0.0043N$

The centripetal force is

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

$m = m a s s = 4 k g$ $m=mass=4kg$

$v = 0.06 m s^{- 1}$ $v=0.06ms^-1$

The variation in centripetal force is

$DeltaF=F_2-F_1$

$F_1=4*0.06^2/0.56=0.0257N$

$F_2=4*0.06^2/0.48=0.03N$

$DeltaF=0.03-0.0257=0.0043N$

A model train with a mass of 4 kg4kg4 kg is moving along a track at 6 (cm)/s6cms6 (cm)/s. If the curvature of the track changes from a radius of 56 cm56cm56 cm to 48 cm48cm48 cm, by how much must the centripetal force applied by the tracks change?