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 $240 c m$ $240 cm$ , by how much must the centripetal force applied by the tracks change?

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Answer 1 · 2017-03-11T09:31:10

The change in centripetal force is $= 0.192 N$ $=0.192N$

The centripetal force is

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

$m = m a s s = 8 k g$ $m=mass=8kg$

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

The variation in centripetal force is

$DeltaF=F_2-F_1$

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

$F_2=8*0.12^2/2.40=0.048N$

$DeltaF=0.24-0.048=0.192N$

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 240 cm240cm240 cm, by how much must the centripetal force applied by the tracks change?