A model train with a mass of #8 kg# is moving along a track at #15 (cm)/s#. If the curvature of the track changes from a radius of #45 cm# to #225 cm#, by how much must the centripetal force applied by the tracks change?

Question

A model train with a mass of #8  kg# is moving along a track at #15 (cm)/s#.  If the curvature of the track changes from a radius of #45 cm# to #225 cm#, by how much must the centripetal force applied by the tracks change?

Charlotte Aaron · Accepted Answer

The centripetal force is #0.1N# The centripetal force is #F=(mv^2)/r# The variation in centripetal force is #DeltaF=(mv^2)/(Delta r)# #=(8*(0.15)^2)/(2.25-0.45)# #=(8*0.0225)/1.8# #=0.1N#

Isabella Alvarez · Answer

undefined The centripetal force must increase by 40 N.

Isabella Alvarez · Answer

undefined The centripetal force required to keep an object moving in a curved path is given by the formula:

$$ F_{\text{centripetal}} = \frac{mv^2}{r} $$

Where:
- $ F_{\text{centripetal}} $ is the centripetal force,
- $ m $ is the mass of the object,
- $ v $ is the velocity of the object, and
- $ r $ is the radius of curvature of the path.

Given:
- Mass $ m = 8 $ kg,
- Initial radius $ r_1 = 45 $ cm,
- Final radius $ r_2 = 225 $ cm, and
- Initial velocity $ v = 15 $ cm/s.

First, we need to convert all measurements to SI units (kilograms, meters, seconds) for consistency in calculations.

$$ r_1 = 45 \, \text{cm} = 0.45 \, \text{m} $$
$$ r_2 = 225 \, \text{cm} = 2.25 \, \text{m} $$
$$ v = 15 \, \text{cm/s} = 0.15 \, \text{m/s} $$

Now we can calculate the initial and final centripetal forces:

$$ F_{\text{initial}} = \frac{m \cdot v^2}{r_1} $$
$$ F_{\text{final}} = \frac{m \cdot v^2}{r_2} $$

Substitute the given values:

$$ F_{\text{initial}} = \frac{8 \times (0.15)^2}{0.45} \approx 0.8 \, \text{N} $$
$$ F_{\text{final}} = \frac{8 \times (0.15)^2}{2.25} \approx 0.0533 \, \text{N} $$

The change in centripetal force is the difference between the final and initial forces:

$$ \Delta F_{\text{centripetal}} = F_{\text{final}} - F_{\text{initial}} $$
$$ \Delta F_{\text{centripetal}} = 0.0533 \, \text{N} - 0.8 \, \text{N} \approx -0.7467 \, \text{N} $$

Therefore, the centripetal force applied by the tracks must decrease by approximately $ 0.7467 \, \text{N} $ to accommodate the change in curvature from a radius of 45 cm to 225 cm.