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 $180 c m$ $180 cm$ to $81 c m$ $81 cm$ , by how much must the centripetal force applied by the tracks change?

Question

1289 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-07-29T05:23:55

The change in centripetal force is $= 0.044 N$ $=0.044N$

The centripetal force is

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

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

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

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

The variation in centripetal force is

$DeltaF=F_2-F_1$

$F_1=mv^2/r_1=8*0.09^2/1.8=0.036N$

$F_2=mv^2/r_2=8*0.09^2/0.81=0.08N$

$DeltaF=0.08-0.036=0.044N$

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 180 cm180cm180 cm to 81 cm81cm81 cm, by how much must the centripetal force applied by the tracks change?