A model train, with a mass of #7 kg#, is moving on a circular track with a radius of #1 m#. If the train's rate of revolution changes from #1/3 Hz# to #2/7 Hz#, by how much will the centripetal force applied by the tracks change by?

Answer 1

The centripetal force changes by #=0.63N#

Centripetal force is what

#F=mr omega^2#
mass is #m=7kg#
radius #r=1m#

Delta omega is equal to (1/3-2/7)*2pi, or (2/21pi)rad^-1.

The centripetal force fluctuation is

#DeltaF=mr(Delta omega)^2#
#=7*1*(2/21pi)^2#
#=0.63N#
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Answer 2

The centripetal force applied by the tracks on the model train is given by the formula:

( F = m \times (2\pi f)^2 \times r )

Where:

  • ( F ) is the centripetal force,
  • ( m ) is the mass of the train (7 kg),
  • ( f ) is the frequency of revolution (in Hz),
  • ( r ) is the radius of the circular track (1 m).

Using the initial and final frequencies, we can calculate the initial and final centripetal forces, and then find the difference:

Initial frequency, ( f_1 = \frac{1}{3} ) Hz Final frequency, ( f_2 = \frac{2}{7} ) Hz

Initial centripetal force, ( F_1 = 7 \times (2\pi \times \frac{1}{3})^2 \times 1 ) Final centripetal force, ( F_2 = 7 \times (2\pi \times \frac{2}{7})^2 \times 1 )

( F_1 = 7 \times (2\pi \times \frac{1}{3})^2 \times 1 ) ( F_1 = 7 \times (2\pi \times \frac{1}{3})^2 ) ( F_1 = 7 \times (2\pi \times \frac{1}{3} \times \frac{1}{3}) ) ( F_1 = 7 \times (2\pi \times \frac{1}{9}) ) ( F_1 = 7 \times (\frac{2\pi}{9}) ) ( F_1 = \frac{14\pi}{9} )

( F_2 = 7 \times (2\pi \times \frac{2}{7})^2 \times 1 ) ( F_2 = 7 \times (2\pi \times \frac{2}{7})^2 ) ( F_2 = 7 \times (2\pi \times \frac{2}{7} \times \frac{2}{7}) ) ( F_2 = 7 \times (2\pi \times \frac{4}{49}) ) ( F_2 = 7 \times (\frac{8\pi}{49}) ) ( F_2 = \frac{56\pi}{49} )

Difference in centripetal force, ( \Delta F = F_2 - F_1 )

( \Delta F = \frac{56\pi}{49} - \frac{14\pi}{9} ) ( \Delta F = \frac{56\pi}{49} - \frac{14\pi}{9} \times \frac{63}{63} ) ( \Delta F = \frac{56\pi}{49} - \frac{98\pi}{63} ) ( \Delta F = \frac{56\pi \times 63}{49 \times 63} - \frac{98\pi \times 49}{63 \times 49} ) ( \Delta F = \frac{3528\pi - 4802\pi}{3087} ) ( \Delta F = \frac{-1274\pi}{3087} )

Therefore, the change in centripetal force applied by the tracks is ( \frac{-1274\pi}{3087} ) or approximately ( -1.30 ) N.

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Answer from HIX Tutor

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.

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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