A model train with a mass of $4 k g$ $4 kg$ is moving along a track at $9 \frac{c m}{s}$ $9 (cm)/s$ . If the curvature of the track changes from a radius of $42 c m$ $42 cm$ to $45 c m$ $45 cm$ , by how much must the centripetal force applied by the tracks change?

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Answer 1 · 2017-09-03T15:31:15

The change in centripetal force is $= 0.005 N$ $=0.005N$

The centripetal force is

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

The mass, $m = (4) k g$ $m=(4)kg$

The speed, $v = (0.09) m s^{- 1}$ $v=(0.09)ms^-1$

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

The variation in centripetal force is

$DeltaF=F_2-F_1$

$F_1=mv^2/r_1=4*0.09^2/0.42=0.077N$

$F_2=mv^2/r_2=4*0.09^2/0.45=0.0722N$

$DeltaF=0.077-0.072=0.005N$

A model train with a mass of 4 kg4kg4 kg is moving along a track at 9 (cm)/s9cms9 (cm)/s. If the curvature of the track changes from a radius of 42 cm42cm42 cm to 45 cm45cm45 cm, by how much must the centripetal force applied by the tracks change?