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 $36 c m$ $36 cm$ to $45 c m$ $45 cm$ , by how much must the centripetal force applied by the tracks change?

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Answer 1 · 2017-03-08T15:59:21

The change in centripeta force is $= 0.008 N$ $=0.008N$

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

$DeltaF=F_2-F_1$

$F_1=4*0.06^2/0.36=0.04N$

$F_2=4*0.06^2/0.45=0.032N$

$DeltaF=0.04-0.032=0.008N$

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 36 cm36cm36 cm to 45 cm45cm45 cm, by how much must the centripetal force applied by the tracks change?