A model train with a mass of $3 k g$ $3 kg$ 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 · 2017-04-06T10:04:03

The change in centripetal force is $= 0.04 N m$ $=0.04Nm$

The centripetal force is

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

mass, $m = 3 k g$ $m=3kg$

speed, $v = 0.08 m s^{- 1}$ $v=0.08ms^-1$

radius, $= r$ $=r$

The variation in centripetal force is

$DeltaF=F_2-F_1$

$F_1=3*0.08^2/0.12=0.16N$

$F_2=3*0.08^2/0.16=0.12N$

$DeltaF=0.16-0.12=0.04N$

A model train with a mass of 3 kg3kg3 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?