A model train with a mass of $8 k g$ $8 kg$ is moving along a track at $16 \frac{c m}{s}$ $16 (cm)/s$ . If the curvature of the track changes from a radius of $48 c m$ $48 cm$ to $120 c m$ $120 cm$ , by how much must the centripetal force applied by the tracks change?

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Answer 1 · 2016-12-18T14:30:55

The centripetal force would decrease from 0.427 N to 0.171 N, a change of 0.256 N

Centripetal force required for a circular path is found from

$F_{c}$ $F_c$ = $\frac{m v^{2}}{r}$ $(mv^2)/r$

First, we should get rid of the centimetres! Convert both to metres to avoid conversion factor errors.

When travelling at $0.16 \frac{m}{s}$ $0.16 m/s$ on the track having $r = 0.48 m$ $r = 0.48m$ , the required force is

$F_{c}$ $F_c$ = $\frac{8 \times {0.16}^{2}}{0.48}$ $(8 xx 0.16^2)/0.48$ = $0.427 N$ $0.427 N$

On the track having the greater radius, less force is required (note the $r$ $r$ in the denominator)

$F_{c}$ $F_c$ = $\frac{8 \times {0.16}^{2}}{1.2}$ $(8 xx 0.16^2)/1.2$ = $0.171 N$ $0.171 N$

So, the change is centripetal force is $0.256 N$ $0.256 N$

A model train with a mass of 8kg8 kg is moving along a track at 16cms16 (cm)/s. If the curvature of the track changes from a radius of 48cm48 cm to 120cm120 cm, by how much must the centripetal force applied by the tracks change?