A model train with a mass of $9 k g$ $9 kg$ is moving along a track at $15 \frac{c m}{s}$ $15 (cm)/s$ . If the curvature of the track changes from a radius of $45 c m$ $45 cm$ to $30 c m$ $30 cm$ , by how much must the centripetal force applied by the tracks change?

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Answer 1 · 2017-07-02T06:30:52

The change in centripetal force is $= 0.225 N$ $=0.225N$

The centripetal force is

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

mass, $m = 9 k g$ $m=9kg$

speed, $v = 0.15 m s^{- 1}$ $v=0.15ms^-1$

radius, $= (r) m$ $=(r) m$

The variation in centripetal force is

$DeltaF=F_2-F_1$

$F_1=mv^2/r_1=9*0.15^2/0.45=0.45N$

$F_2=mv^2/r_2=9*0.15^2/0.30=0.675N$

$DeltaF=0.675-0.45=0.225N$

A model train with a mass of 9 kg9kg9 kg is moving along a track at 15 (cm)/s15cms15 (cm)/s. If the curvature of the track changes from a radius of 45 cm45cm45 cm to 30 cm30cm30 cm, by how much must the centripetal force applied by the tracks change?