A model train with a mass of $6 k g$ $6 kg$ is moving along a track at $8 \frac{c m}{s}$ $8 (cm)/s$ . If the curvature of the track changes from a radius of $12 c m$ $12 cm$ to $16 c m$ $16 cm$ , by how much must the centripetal force applied by the tracks change?

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Answer 1 · 2017-12-25T16:17:52

The change in centripetal force is $= 0.08 N$ $=0.08N$

The centripetal force is

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

The mass is of the train $m = 6 k g$ $m=6kg$

The velocity is $v = 0.08 m s^{- 1}$ $v=0.08ms^-1$

The radii are

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

and

$r_{2} = 0.16 m$ $r_2=0.16m$

The variation in the centripetal force is

$DeltaF=F_2-F_1$

The centripetal forces are

$F_1=6*0.08^2/0.12=0.32N$

$F_2=6*0.08^2/0.16=0.24N$

$DeltaF=0.32-0.24=0.08N$

A model train with a mass of 6 kg6kg6 kg is moving along a track at 8 (cm)/s8cms8 (cm)/s. If the curvature of the track changes from a radius of 12 cm12cm12 cm to 16 cm16cm16 cm, by how much must the centripetal force applied by the tracks change?