A solid flywheel of 20 kg mass and 120 mm radius revolves at `600 "revmin"^(-1)`. With what force must a brake lining be pressed against it for the flywheel to stop in 3s, if the coefficient of friction is 0.1 ?

Explanation

To solve the problem of how much force must be applied to the brake lining of a flywheel to stop it in a given time, we can follow these steps:

### Step 1: Convert the initial angular velocity from RPM to radians per second
The initial angular velocity \( \omega_0 \) is given as 600 revolutions per minute (RPM). We need to convert this to radians per second.

\[ \omega_0 = 600 \, \text{rev/min} \times \frac{2\pi \, \text{rad}}{1 \, \text{rev}} \times \frac{1 \, \text{min}}{60 \, \text{s}} = 62.83 \, \text{rad/s} \]
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