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 $32 c m$ $32 cm$ to $45 c m$ $45 cm$ , by how much must the centripetal force applied by the tracks change?

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Answer 1 · 2016-04-17T09:53:24

$- 0.26325 N$ $-0.26325N$

Centripetal (center-seeking) force is given by the equation

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

where $m$ $m$ is mass, $v$ $v$ is velocity and $r$ $r$ is radius.

At the beginning, when the radius is $0.32 m$ $0.32m$ and velocity $0.18 m s^{- 1}$ $0.18ms^-1$ , the centripetal force would be

$F = \frac{9 k g \cdot 0.0324 m^{2} s^{- 2}}{0.32 m}$ $F=(9kg*0.0324m^2s^-2)/(0.32m)$
$0.91125 N$ $0.91125N$

In the second part, when the radius has expanded to $0.45 m$ $0.45m$ , centripetal force is

$F = \frac{9 k g \cdot 0.0324 m^{2} s^{- 2}}{0.45 m}$ $F=(9kg*0.0324m^2s^-2)/(0.45m)$
$= 0.648 N$ $=0.648N$

The change in centripetal force is then found by simple subtraction,

$DeltaF=0.648N-0.91125N$
$=-0.26325N$

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 32 cm32cm32 cm to 45 cm45cm45 cm, by how much must the centripetal force applied by the tracks change?