## Related questions with answers

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?

Solutions

Verified*$\large{\text{PROBLEM:}}$*

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

*$\large{\text{GIVEN:}}$*

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

The braking torque will be equal to the friction force times the radius. Let us remember that the friction force is given as just the product of the coefficient of friction and the normal force. Writing that, we have

$\tau =F_fR=\mu NR,$

from where we have our force as

$N=\frac{\tau}{\mu R}.$

We thus need to find the torque.

Let us remember a torque is connected to the moment of inertia $I$ and the angular acceleration it will cause as

$\tau =I\alpha .$

Substituting this, we have

$N=\frac{I\alpha}{\mu R}.$

We don't know the angular acceleration, however. Let's remember that since the acceleration is constant, the following holds true:

$\omega^2 =2\alpha \phi,$

from where we can find the angular acceleration as

$\alpha =\frac{\omega^2}{2\phi }.$

Substituting, we get

$N=\frac{I\omega^2}{2\mu \phi R}.$

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