A model train, with a mass of $12 k g$ $12 kg$ , is moving on a circular track with a radius of $9 m$ $9 m$ . If the train's rate of revolution changes from $6 H z$ $6 Hz$ to $8 H z$ $8 Hz$ , by how much will the centripetal force applied by the tracks change by?

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Answer 1 · 2016-03-11T17:38:26

The change in Centripetal Force will be $48 N \to 85.3 N$ $48 N to 85.3 N$ .

Here:

$m = 12$ $m = 12$
$v_{1} = 6$ $v_1 = 6$
$v_{2} = 8$ $v_2 = 8$
$r = 9$ $r = 9$

We know:

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

For first rate of revolution. i.e. $6 H z$ $6 Hz$

$C F_{1} = \frac{12 \cdot 6^{2}}{9}$ $CF_1 = (12 *6^2)/9$

$C F_{1} = \frac{12 \cdot 36}{9}$ $CF_1 = (12 * 36)/9$

$C F_{1} = \frac{432}{9}$ $CF_1 = (432)/9$

$C F_{1} = 48 N$ $CF_1 = 48 N$

For second rate of revolution. i.e. $8 H z$ $8 Hz$

$C F_{2} = \frac{12 \cdot 8^{2}}{9}$ $CF_2 = (12 *8^2)/9$

$C F_{2} = \frac{12 \cdot 64}{9}$ $CF_2 = (12 * 64)/9$

$C F_{2} = \frac{768}{9}$ $CF_2 = (768)/9$

$C F_{2} = 85.3 N$ $CF_2 = 85.3 N$

The change in Centripetal Force will be $48 N \to 85.3 N$ $48 N to 85.3 N$ .

A model train, with a mass of 12 kg12kg12 kg, is moving on a circular track with a radius of 9 m9m9 m. If the train's rate of revolution changes from 6 Hz6Hz6 Hz to 8 Hz8Hz8 Hz, by how much will the centripetal force applied by the tracks change by?