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A solid, uniform cylinder with mass 8.25 $\mathrm{kg}$ and diameter 15.0 $\mathrm{cm}$ is spinning at 220 $\mathrm{rpm}$ 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 $0.333 .$ What must the applied normal force be to bring the cylinder to rest after it has turned through 5.25 revolutions?

7.47 $\mathrm{N}$

Physics 101 Mechanics

Chapter 10

Dynamics of Rotational Motion

Newton's Laws of Motion

Rotation of Rigid Bodies

Equilibrium and Elasticity

Cornell University

University of Washington

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So we want to find the applied normal force, so M equals eight point to five kilograms. Ah, the diameter is 15 centimeters. We can immediately say that the radius is going to be equal to a 0.75 It's 0.75 meters. We know that the initial angular velocity is 22 other 220 revolutions per minute. Ah, this will be times two pi radiance Pere revolution and then times one minute for every 60 seconds. And this is giving us 23 0.0 for radiance per second. We know that the coefficient of connect fiction is 0.33 rather 0.333 And we know that the moments of inertia here is going to be equal to half em are squared splices calculate this 1/2 times 8.25 times 0.75 squared. And this is giving us a point zero 232 kilogram meters squared. At this point, we know that Delta Theta also equals 5.25 revolutions. And we're going to say that this is going to be 10.5 pie radiance again. There are two pyre. Etienne's for everyone revolution. So we first need to find the angular acceleration we can use. Angular Kenna Matics. So Omega Final squared equals omega initial squared plus two times Alfa Times, Delta theta. We know that that this is coming to arrest so Alfa will simply be equal to negative Omega initial squared divided by two times Delta theta. And this is equaling negative 23.4 squared, divided by two times 10.5, ten 10.5 pie. And this is giving us negative 8.46 radiance per second squared. Now, at this point, we can find the normal force by using the sum of torque so that some of torques is going to be equal to the torque of friction. And this will be quick to negative negative force of friction times the radius. This will be equal to negative mu k f and times are, and we know that the torque the sum of the torque also equals the moment of inertia, times the anger, angular acceleration. So to find the force normal, this is simply going to be equal to negative negative one times the moment of inertia times the angular, angular acceleration divided by mu K Times the radius. And this will be negative 0.0 232 times negative, 8.0 for six, divided by ah 0.333 times the radius of 0.75 And we're getting that. The force normal is going to be equal to 7.47 Newton's. So this will be our final answer. That is the end of the solution. Thank you for watching.

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