## College Physics (4th Edition)

The normal force applied to a wheel by one of the brake pads is $~1.20~N$
We can find the initial angular speed of each wheel: $\omega_0 = \frac{v_0}{r} = \frac{7.5~m/s}{0.35~m} = 21.43~rad/s$ We can find the rate of angular deceleration of each wheel as the bicycle comes to a stop: $\omega_f = \omega_0+\alpha~t$ $\alpha = \frac{\omega_f - \omega_0}{t}$ $\alpha = \frac{0 - 21.43~rad/s}{4.5~s}$ $\alpha = -4.76~rad/s^2$ Since there are four brake pads in total and there are two wheels, there are two brake pads for each wheel. We can use the magnitude of the angular acceleration to find the normal force $F_N$ applied to the wheel by each brake pad. We can consider the normal force of two brake pads on one wheel: $\tau = I~\alpha$ $r~F_N~\mu_k+r~F_N~\mu_k = Mr^2~\alpha$ $2~r~F_N~\mu_k = Mr^2~\alpha$ $F_N = \frac{Mr~\alpha}{2~\mu_k}$ $F_N = \frac{(1.3~kg)(0.35~m)(4.76~rad/s^2)}{(2)(0.90)}$ $F_N = 1.20~N$ The normal force applied to a wheel by one of the brake pads is $~1.20~N$.