A model train with a mass of $9 k g$ $9 kg$ 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 $36 c m$ $36 cm$ to $42 c m$ $42 cm$ , by how much must the centripetal force applied by the tracks change?

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Answer 1 · 2018-08-01T14:03:42

The change in centripetal force is $= 0.12 N$ $=0.12N$

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 $m = 9 k g$ $m=9kg$

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

The radii of the tracks are

$r_{1} = 0.36 m$ $r_1=0.36m$

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

$||F_1||=9*0.18^2/0.36=0.81N$

$||F_2||=9*0.18^2/0.42=0.69N$

$DeltaF=|F_1-F_2|=0.81-0.69=0.12N$

A model train with a mass of 9 kg9kg9 kg is moving along a track at 18 (cm)/s18cms18 (cm)/s. If the curvature of the track changes from a radius of 36 cm36cm36 cm to 42 cm42cm42 cm, by how much must the centripetal force applied by the tracks change?