A model train with a mass of $8 k g$ $8 kg$ is moving along a track at $9 \frac{c m}{s}$ $9 (cm)/s$ . If the curvature of the track changes from a radius of $27 c m$ $27 cm$ to $72 c m$ $72 cm$ , by how much must the centripetal force applied by the tracks change?

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Answer 1 · 2017-11-07T06:38:03

The change in centripeta force is $= 0.15 N$ $=0.15N$

The centripetal force is

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

The mass, $m = (8) k g$ $m=(8)kg$

The speed, $v = (0.09) m s^{- 1}$ $v=(0.09)ms^-1$

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

The variation in centripetal force is

$DeltaF=F_2-F_1$

$F_1=mv^2/r_1=8*0.09^2/0.27=0.24N$

$F_2=mv^2/r_2=8*0.09^2/0.72=0.09N$

$DeltaF=0.24-0.09=0.15N$

A model train with a mass of 8 kg8kg8 kg is moving along a track at 9 (cm)/s9cms9 (cm)/s. If the curvature of the track changes from a radius of 27 cm27cm27 cm to 72 cm72cm72 cm, by how much must the centripetal force applied by the tracks change?