A model train, with a mass of #8 kg#, is moving on a circular track with a radius of #6 m#. If the train's rate of revolution changes from #3 Hz# to #2 Hz#, by how much will the centripetal force applied by the tracks change by?

Question

A model train, with a mass of #8 kg#, is moving on a circular track with a radius of #6 m#.  If the train's rate of revolution changes from #3 Hz# to #2 Hz#, by how much will the centripetal force applied by the tracks change by?

Lily Adams · Accepted Answer

The centripetal force is #=1895N# The centripetal force is #F=mr omega^2# The variation in centripetal force is #DeltaF=mr(Delta omega)^2# #=8*6*((3-2)2pi)^2# #=8*6*4*pi^2# #=1895N#

Charles Aeschliman · Answer

undefined The change in centripetal force applied by the tracks can be calculated using the formula:

ΔF = mω²(Δr)

Where:
- ΔF is the change in centripetal force,
- m is the mass of the train (8 kg),
- ω is the angular velocity (2πf, where f is the frequency),
- Δr is the change in radius (0, as the radius remains constant).

Given that the angular velocity changes from 2π(3 Hz) to 2π(2 Hz), the change in frequency (Δf) is 3 Hz - 2 Hz = 1 Hz.

Using the formula for angular velocity (ω = 2πf), the initial angular velocity (ω₁) is 2π(3 Hz) = 6π rad/s, and the final angular velocity (ω₂) is 2π(2 Hz) = 4π rad/s.

Now, we can calculate the change in centripetal force:

ΔF = (8 kg)(6π rad/s)² - (8 kg)(4π rad/s)²

ΔF = 288π² - 128π²

ΔF = 160π² ≈ 5020.53 N

Therefore, the centripetal force applied by the tracks changes by approximately 5020.53 N.

Charles Aeschliman · Answer

undefined The centripetal force applied by the tracks will change by $ 29.4 \, \text{N} $.