A model train with a mass of $4 k g$ $4 kg$ 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 $54 c m$ $54 cm$ to $27 c m$ $27 cm$ , by how much must the centripetal force applied by the tracks change?

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Answer 1 · 2017-03-04T20:33:25

$0.0474 N$ $0.0474N$

Firstly let's just put everything in SI units - change the $c m$ $cm$ to $m$ $m$ to make everything easier to solve.

$8 c m s^{- 1} = 0.08 m s^{- 1}$ $8cms^-1 = 0.08ms^-1$
$54 c m = 0.54 m$ $54cm = 0.54m$
$27 = 0.27 m$ $27 = 0.27m$

Now, the equation for centripetal force is

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

where $F$ $F$ is force, $a$ $a$ is acceleration, $m$ $m$ is mass $,$ $,$ v $i s v e l o c i t y$ $is velocity$ and $r$ $r$ is radius.

In this case, the radius is changing, so we can say that

$DeltaF = (mv^2)/r_2 - (mv^2)/r_1$

Putting in the values that we know,

$DeltaF = (4*0.08^2)/0.27 - (4*0.08^2)/0.54$

$= 0.0474N$

A model train with a mass of 4 kg4kg4 kg is moving along a track at 8 (cm)/s8cms8 (cm)/s. If the curvature of the track changes from a radius of 54 cm54cm54 cm to 27 cm27cm27 cm, by how much must the centripetal force applied by the tracks change?