# A model train, with a mass of #2 kg#, is moving on a circular track with a radius of #4 m#. If the train's rate of revolution changes from #1/2 Hz# to #2/8 Hz#, by how much will the centripetal force applied by the tracks change by?

The centripetal force change by

Centripetal force is what

The centripetal force fluctuation is

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To calculate the change in centripetal force, we can use the formula:

[ F = m \cdot \omega^2 \cdot r ]

Where:

- ( F ) is the centripetal force,
- ( m ) is the mass of the train (2 kg),
- ( \omega ) is the angular velocity (given by ( 2\pi f ), where ( f ) is the frequency),
- ( r ) is the radius of the circular track (4 m).

Initially: [ f_1 = \frac{1}{2} \text{ Hz} \Rightarrow \omega_1 = 2\pi f_1 ] [ f_2 = \frac{2}{8} \text{ Hz} \Rightarrow \omega_2 = 2\pi f_2 ]

Then, we calculate the initial and final centripetal forces:

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

Finally, we find the change in centripetal force: [ \Delta F = F_2 - F_1 ]

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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