A model train with a mass of $1 k g$ $1 kg$ 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 $12 c m$ $12 cm$ to $8 c m$ $8 cm$ , by how much must the centripetal force applied by the tracks change?

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Answer 1 · 2018-01-05T06:29:02

The change in centripetal force is $= 0.015 N$ $=0.015N$

The centripetal force is

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

The mass is of the train $m = 1 k g$ $m=1kg$

The velocity of the train is $v = 0.06 m s^{- 1}$ $v=0.06ms^-1$

The radii are

$r_{1} = 0.12 m$ $r_1=0.12m$

and

$r_{2} = 0.08 m$ $r_2=0.08m$

The variation in the centripetal force is

$DeltaF=F_2-F_1$

The centripetal forces are

$F_1=1*0.06^2/0.12=0.03N$

$F_2=1*0.06^2/0.08=0.045N$

$DeltaF=0.045-0.03=0.015N$

A model train with a mass of 1 kg1kg1 kg is moving along a track at 6 (cm)/s6cms6 (cm)/s. If the curvature of the track changes from a radius of 12 cm12cm12 cm to 8 cm8cm8 cm, by how much must the centripetal force applied by the tracks change?