A model train with a mass of $2 k g$ $2 kg$ 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 $5 c m$ $5 cm$ to $25 c m$ $25 cm$ , by how much must the centripetal force applied by the tracks change?

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Answer 1 · 2018-03-04T07:43:24

The change in centripetal force is $= 0.277 N$ $=0.277N$

The centripetal force is

${\to F}_{C} = \frac{m v^{2}}{r} \cdot \to r$ $vecF_C=(mv^2)/r*vecr$

The mass is of the train $m = 2 k g$ $m=2kg$

The velocity of the train is $v = 0.09 m s^{- 1}$ $v=0.09ms^-1$

The radii of the tracks are

$r_{1} = 0.05 m$ $r_1=0.05m$

and

$r_{2} = 0.25 m$ $r_2=0.25m$

The variation in the centripetal force is

$DeltaF=F_2-F_1$

The centripetal forces are

$||F_1||=2*0.09^2/0.05=0.324N$

$||F_2||=2*0.09^2/0.25=0.0648N$

$DeltaF=F_1-F_2=0.342-0.0648=0.277N$

A model train with a mass of 2 kg2kg2 kg is moving along a track at 9 (cm)/s9cms9 (cm)/s. If the curvature of the track changes from a radius of 5 cm5cm5 cm to 25 cm25cm25 cm, by how much must the centripetal force applied by the tracks change?