A model train, with a mass of #5 kg#, is moving on a circular track with a radius of #9 m#. If the train's rate of revolution changes from #4 Hz# to #5 Hz#, by how much will the centripetal force applied by the tracks change by?

I think the best way to do this is to figure out how the time period of rotation changes:

Period and frequency are each other's reciprocal:

So the time period of rotation of the train changes from 0.25 seconds to 0.2 seconds. When the frequency increases. (We have more rotations per second)

However, the train still has to cover the full distance of the circumference of the circular track.

Speed=distance/time

Then we can find the centripetal force in both scenarios:

So, when the frequency is 4 Hz:

To calculate the change in centripetal force, we use the formula: ( F = m \cdot \omega^2 \cdot r ), where:

• ( F ) is the centripetal force,
• ( m ) is the mass of the object (train),
• ( \omega ) is the angular velocity, and
• ( r ) is the radius of the circular track.

Given:

• ( m = 5 ) kg,
• ( r = 9 ) m,
• ( \omega_1 = 2\pi \cdot 4 ) (initial angular velocity),
• ( \omega_2 = 2\pi \cdot 5 ) (final angular velocity).

We calculate the initial centripetal force (( F_1 )) and the final centripetal force (( F_2 )), then find the difference:

[ F_1 = m \cdot \omega_1^2 \cdot r ] [ F_2 = m \cdot \omega_2^2 \cdot r ]

[ F_1 = 5 \cdot (2\pi \cdot 4)^2 \cdot 9 ] [ F_2 = 5 \cdot (2\pi \cdot 5)^2 \cdot 9 ]

[ F_1 \approx 2261.946 , \text{N} ] [ F_2 \approx 2827.432 , \text{N} ]

The change in centripetal force (( \Delta F )) is:

[ \Delta F = F_2 - F_1 ] [ \Delta F \approx 565.486 , \text{N} ]

Therefore, the centripetal force applied by the tracks changes by approximately ( 565.486 , \text{N} ).