A model train, with a mass of #4 kg#, is moving on a circular track with a radius of #4 m#. If the train's rate of revolution changes from #1/9 Hz# to #1/3 Hz#, by how much will the centripetal force applied by the tracks change by?

Centripetal force is what

The frequencies that are

The centripetal force fluctuation is

The centripetal force acting on an object moving in a circular path is given by the formula:

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

Where:

• ( F ) is the centripetal force
• ( m ) is the mass of the object
• ( \omega ) is the angular velocity of the object (in radians per second)
• ( r ) is the radius of the circular path

The angular velocity ( \omega ) is related to the frequency ( f ) by the formula ( \omega = 2\pi f ).

Given:

• ( m = 4 , \text{kg} )
• Initial frequency ( f_1 = \frac{1}{9} , \text{Hz} )
• Final frequency ( f_2 = \frac{1}{3} , \text{Hz} )
• ( r = 4 , \text{m} )

First, we need to find the initial and final angular velocities:

[ \omega_1 = 2\pi f_1 ] [ \omega_1 = 2\pi \times \frac{1}{9} ] [ \omega_1 = \frac{2\pi}{9} ]

[ \omega_2 = 2\pi f_2 ] [ \omega_2 = 2\pi \times \frac{1}{3} ] [ \omega_2 = \frac{2\pi}{3} ]

Now, we can calculate the initial and final centripetal forces:

[ F_1 = m \times \omega_1^2 \times r ] [ F_1 = 4 \times \left(\frac{2\pi}{9}\right)^2 \times 4 ]

[ F_2 = m \times \omega_2^2 \times r ] [ F_2 = 4 \times \left(\frac{2\pi}{3}\right)^2 \times 4 ]

Finally, we can find the change in centripetal force:

[ \Delta F = F_2 - F_1 ]

Substitute the calculated values and compute:

[ \Delta F = \left(4 \times \left(\frac{2\pi}{3}\right)^2 \times 4\right) - \left(4 \times \left(\frac{2\pi}{9}\right)^2 \times 4\right) ]

[ \Delta F ≈ 42.97 , \text{N} ]

Therefore, the centripetal force applied by the tracks changes by approximately ( 42.97 , \text{N} ).