A model train with a mass of $4 k g$ $4 kg$ is moving along a track at $18 \frac{c m}{s}$ $18 (cm)/s$ . If the curvature of the track changes from a radius of $84 c m$ $84 cm$ to $42 c m$ $42 cm$ , by how much must the centripetal force applied by the tracks change?

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Answer 1 · 2018-03-22T08:42:34

The change in centripetal force is $= 0.155 N$ $=0.155N$

The centripetal force is

${\to F}_{C} = \frac{m v^{2}}{r} \cdot \to r$ $vecF_C=(mv^2)/r*vecr$

The mass is of the train $m = 4 k g$ $m=4kg$

The velocity of the train is $v = 0.18 m s^{- 1}$ $v=0.18ms^-1$

The radii of the tracks are

$r_{1} = 0.84 m$ $r_1=0.84m$

and

$r_{2} = 0.42 m$ $r_2=0.42m$

The variation in the centripetal force is

$DeltaF=F_2-F_1$

The centripetal forces are

$||F_1||=4*0.18^2/0.84=0.154N$

$||F_2||=4*0.18^2/0.42=0.309N$

$DeltaF=F_1-F_2=0.154-0.309=-0.155N$

A model train with a mass of 4 kg4kg4 kg is moving along a track at 18 (cm)/s18cms18 (cm)/s. If the curvature of the track changes from a radius of 84 cm84cm84 cm to 42 cm42cm42 cm, by how much must the centripetal force applied by the tracks change?