A model train with a mass of $6 k g$ $6 kg$ 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 $3 c m$ $3 cm$ to $15 c m$ $15 cm$ , by how much must the centripetal force applied by the tracks change?

Question

1390 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-11-18T07:50:55

The change in centripetal force is $= 1.024 N$ $=1.024N$

The centripetal force is

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

The mass, $m = (6) k g$ $m=(6)kg$

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

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

$r_{1} = 0.03 m$ $r_1=0.03m$

$r_{2} = 0.15 m$ $r_2=0.15m$

The variation in centripetal force is

$DeltaF=F_2-F_1$

$F_1=mv^2/r_1=6*0.08^2/0.03=1.28N$

$F_2=mv^2/r_2=6*0.08^2/0.15=0.256N$

$DeltaF=F_1-F_2=1.28-0.256=1.024N$

A model train with a mass of 6 kg6kg6 kg is moving along a track at 8 (cm)/s8cms8 (cm)/s. If the curvature of the track changes from a radius of 3 cm3cm3 cm to 15 cm15cm15 cm, by how much must the centripetal force applied by the tracks change?