A model train with a mass of #9 kg# is moving along a track at #15 (cm)/s#. If the curvature of the track changes from a radius of #46 cm# to #35 cm#, by how much must the centripetal force applied by the tracks change?

Question

A model train with a mass of #9  kg# is moving along a track at #15 (cm)/s#.  If the curvature of the track changes from a radius of #46 cm# to #35 cm#, by how much must the centripetal force applied by the tracks change?

Julia Aguilar · Accepted Answer

The variation in centripetal force is #=0.14N#

The centripetal force is

#F=(mv^2)/r#

#m=mass=9kg#

#v=0.15ms^-1#

The variation in centripetal force is

#DeltaF=F_2-F_1#

#F_1=9*0.15^2/0.46=0.44N#

#F_2=9*0.15^5/0.35=0.58N#

#DeltaF=0.58-0.44=0.14N#

William Audy · Answer

undefined To find the change in centripetal force, you can use the formula:

$$ \Delta F = \frac{{mv^2}}{{r_1}} - \frac{{mv^2}}{{r_2}} $$

Where:
- $ \Delta F $ is the change in centripetal force,
- $ m $ is the mass of the train (9 kg),
- $ v $ is the velocity of the train (15 cm/s),
- $ r_1 $ is the initial radius (46 cm), and
- $ r_2 $ is the final radius (35 cm).

Plugging in the values:

$$ \Delta F = \frac{{9 \times (15^2)}}{{46}} - \frac{{9 \times (15^2)}}{{35}} $$

$$ \Delta F = \frac{{9 \times 225}}{{46}} - \frac{{9 \times 225}}{{35}} $$

$$ \Delta F = \frac{{2025}}{{46}} - \frac{{2025}}{{35}} $$

$$ \Delta F ≈ 390.67 - 579.64 $$

$$ \Delta F ≈ -188.97 \text{ N} $$

The change in centripetal force is approximately -188.97 N.