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 $25 c m$ $25 cm$ to $15 c m$ $15 cm$ , by how much must the centripetal force applied by the tracks change?

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Answer 1 · 2018-02-09T05:31:04

The change in centrpetal force is $= 0.0384 N$ $=0.0384N$

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 $v = 0.06 m s^{- 1}$ $v=0.06ms^-1$

The radii of the tracks are

$r_{1} = 0.25 m$ $r_1=0.25m$

and

$r_{2} = 0.15 m$ $r_2=0.15m$

The variation in the centripetal force is

$DeltaF=F_2-F_1$

The centripetal forces are

$||F_1||=4*0.06^2/0.25=0.0576N$

$||F_2||=4*0.06^2/0.15=0.096N$

$DeltaF=F_2-F_1=0.096-0.0576=0.0384N$

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 25 cm25cm25 cm to 15 cm15cm15 cm, by how much must the centripetal force applied by the tracks change?