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 $42 c m$ $42 cm$ to $45 c m$ $45 cm$ , by how much must the centripetal force applied by the tracks change?

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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 $42 c m$ $42 cm$ to $45 c m$ $45 cm$ , by how much must the centripetal force applied by the tracks change?

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Answer 1 · 2016-04-23T03:44:22

The mass of the toy train $m = 4 k g$ $m =4kg$
The Velocity of the toy train $v = 21 c m s^{- 1} = 0.21 m s^{- 1}$ $v =21cms^-1=0.21ms^-1$
The radius of first track = $r_{1} = 42 c m = 0.42 m$ $r_1=42cm =0.42m$
The radius of second track = $r_{2} = 45 c m = 0.45 m$ $r_2=45cm =0.45m$
So decrease in centripetal force for changing track is
$= m v^{2} (\frac{1}{r_{1}} - \frac{1}{r_{2}}) = 4 \times {(0.21)}^{2} \times (\frac{1}{0.42} - \frac{1}{0.45}) N \approx .028 N$ $=mv^2(1/r_1-1/r_2)=4xx(0.21)^2xx(1/0.42-1/0.45)N~~.028N$

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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 $42 c m$ $42 cm$ to $45 c m$ $45 cm$ , by how much must the centripetal force applied by the tracks change?

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A model train with a mass of 4kg4 kg is moving along a track at 21cms21 (cm)/s. If the curvature of the track changes from a radius of 42cm42 cm to 45cm45 cm, by how much must the centripetal force applied by the tracks change?

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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 $42 c m$ $42 cm$ to $45 c m$ $45 cm$ , by how much must the centripetal force applied by the tracks change?