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 #5 Hz# to #9 Hz#, by how much will the centripetal force applied by the tracks change by?

Question

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 #5 Hz# to #9 Hz#, by how much will the centripetal force applied by the tracks change by?

Isabella Alvarez · Accepted Answer

The centripetal force changes by #=28.424kN#

The centripetal force is

#F=mr omega^2#

The mass is #m=5kg#

The radius #r=9m#

The change in angular velocity is

#Delta omega=(9-5)*2pi=(8pi)rads^-1#

The variation in centripetal force is

#DeltaF=mr(Delta omega)^2#

#=5*9*(8pi)^2#

#=28424N#

Xavier Adame · Answer

undefined To calculate the change in centripetal force, we'll use the formula:

$$ F = m \times r \times \omega^2 $$

Where:
- $ F $ is the centripetal force
- $ m $ is the mass of the train (5 kg)
- $ r $ is the radius of the circular track (9 m)
- $ \omega $ is the angular velocity

First, we'll calculate the centripetal force with the initial angular velocity:

$$ F_1 = 5 \times 9 \times (2\pi \times 5)^2 $$

Then, we'll calculate the centripetal force with the final angular velocity:

$$ F_2 = 5 \times 9 \times (2\pi \times 9)^2 $$

Finally, we'll find the difference between the two forces:

$$ \Delta F = F_2 - F_1 $$

$$ \Delta F = (5 \times 9 \times (2\pi \times 9)^2) - (5 \times 9 \times (2\pi \times 5)^2) $$