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A solid uniform cylinder with mass 8.25 kg8.25 \mathrm{~kg} and diameter 15.0 cm15.0 \mathrm{~cm} is spinning at 220 rpm220 \mathrm{~rpm} on a thin, frictionless axle that passes along the cylinder axis. You design a simple friction brake to stop the cylinder by pressing the brake against the outer rim with a normal force. The coefficient of kinetic friction between the brake and rim is 0.3330.333. What must the applied normal force be to bring the cylinder to rest after it has turned through 5.255.25 revolutions?

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PROBLEM:\large{\text{PROBLEM:}}

To find the applied normal force to bring the cylinder to rest after it has turned through 5.25 revolutions.

GIVEN:\large{\text{GIVEN:}}

  • Mass of the disk: m=8.25  kgm = 8.25\;\text{kg}
  • Radius of the disk: R=0.0750  mR = 0.0750\;\text{m}
  • Initial angular speed: ω0=0\omega_0 = 0
  • Final angular speed: ω=220  rpm\omega= 220\;\text{rpm}
  • Number of revolutions: θθ0=5.25  rev\theta - \theta_0 = 5.25\;\text{rev}
  • Coefficient of friction: 0.333

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