A model train with a mass of $2 k g$ $2 kg$ is moving along a track at $36 \frac{c m}{s}$ $36 (cm)/s$ . If the curvature of the track changes from a radius of $6 c m$ $6 cm$ to $3 c m$ $3 cm$ , by how much must the centripetal force applied by the tracks change?

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Answer 1 · 2017-03-05T15:42:36

$4.32 N$ $4.32N$

Put the values we have into SI units:

$36 c m s^{- 1} = 0.36 m s^{- 1}$ $36cms^-1 = 0.36ms^-1$
$6 c m = 0.06 m$ $6cm = 0.06m$
$3 c m = 0.03 m$ $3cm = 0.03m$

The equation for centripetal force is

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

so the change in centripetal force is

$DeltaF = (mv^2)/r_2 - (mv^2)/r_1$

Plug in the values we are given,

$DeltaF = (2 xx 0.36^2)/0.03 - (2 xx 0.36^2)/0.06$

$= 8.64 - 4.32 = 4.32N$

A model train with a mass of 2 kg2kg2 kg is moving along a track at 36 (cm)/s36cms36 (cm)/s. If the curvature of the track changes from a radius of 6 cm6cm6 cm to 3 cm3cm3 cm, by how much must the centripetal force applied by the tracks change?