Question

A solid, uniform cylinder with mass 8.15 kg and diameter 15.0 cm is spinning with angular velocity 240 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 0.330.
What must the applied normal force be to bring the cylinder to rest after it has turned through 5.20 revolutions?

Answer 1

Answer:

Normal force will be equal to 8.945 N

Explanation:

We have given mass of the cylinder m = 8.15 kg

Diameter d = 15 cm

So radius $r=\frac{d}{2}=\frac{15}{2}=7.5cm = 0.075 m$

Initial angular velocity $\omega _i=240rpm=\frac{2\times 3.14\times 240}{60}=25.12rad/sec$

As the cylinder finally comes to rest so final angular velocity $\omega _f=0rad/sec$

Before coming to rest cylinder covers a distance of $\Theta =5.20revolution=5.20\times 2\times 3.14=32.656rad$

From third equation of motion $\omega _f^2=\omega_i^2+2\alpha \Theta$

$0^2=25.12^2-2\times \alpha \times32.656$

$\alpha =9.66rad/sec^2$

Coefficient of kinetic friction $\mu _k=0.33$

Moment of inertia of the solid cylinder $I=\frac{1}{2}mr^2$

We know that $\tau =I\alpha$

$F\times r =(\frac{1}{2}mr^2)\times \alpha$

$F =(\frac{1}{2}mr)\times \alpha$

$F=\frac{1}{2}\times 8.15\times 0.075\times 9.66=2.952N$

So normal force will be equal to $N=\frac{F}{\mu _k}=\frac{2.952}{0.33}=8.945N$