# Chapter 12 - Rotation of a Rigid Body - Exercises and Problems: 71

The brake must apply a friction force of 1.6 N

#### Work Step by Step

We can find the moment of inertia of the disk. $I = \frac{1}{2}MR^2$ $I = \frac{1}{2}(2.0~kg)(0.15~m)^2$ $I = 0.0225~kg~m^2$ We can convert the initial angular velocity to units of rad/s $\omega_0 = (300~rpm)(\frac{2\pi~rad}{1~rev})(\frac{1~min}{60~s})$ $\omega_0 = 31.42~rad/s$ We can find the required angular acceleration for the disk to stop in 3.0 seconds. $\alpha = \frac{\omega_f-\omega_0}{t}$ $\alpha = \frac{0-31.42~rad/s}{3.0~s}$ $\alpha = -10.47~rad/s^2$ We can find use the magnitude of the angular acceleration to find the required force of friction. $\tau = I~\alpha$ $(R)~(F_f) = I~\alpha$ $F_f = \frac{I~\alpha}{R}$ $F_f = \frac{(0.0225~kg~m^2)(10.47~rad/s^2)}{0.15~m}$ $F_f = 1.6~N$ The brake must apply a friction force of 1.6 N

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