A model train with a mass of $4 k g$ $4 kg$ is moving along a track at $21 \frac{c m}{s}$ $21 (cm)/s$ . If the curvature of the track changes from a radius of $84 c m$ $84 cm$ to $70 c m$ $70 cm$ , by how much must the centripetal force applied by the tracks change?

Question

1402 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-08-30T05:22:40

The change in centripetal force is $= 0.042 N$ $=0.042N$

The centripetal force is

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

The mass, $m = (4) k g$ $m=(4)kg$

The speed, $v = (0.21) m s^{- 1}$ $v=(0.21)ms^-1$

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

The variation in centripetal force is

$DeltaF=F_2-F_1$

$F_1=mv^2/r_1=4*0.21^2/0.84=0.21N$

$F_2=mv^2/r_2=4*0.21^2/0.70=0.252N$

$DeltaF=0.252-0.21=0.042N$

A model train with a mass of 4 kg4kg4 kg is moving along a track at 21 (cm)/s21cms21 (cm)/s. If the curvature of the track changes from a radius of 84 cm84cm84 cm to 70 cm70cm70 cm, by how much must the centripetal force applied by the tracks change?