A wheel with \(c=\frac{4}{9}\), a mass of \(40.0 \mathrm{~kg}\), and a rim radius of \(30.0 \mathrm{~cm}\) is mounted vertically on a horizontal axis. A 2.00 -kg mass is suspended from the wheel by a rope wound around the rim. Find the angular acceleration of the wheel when the mass is released.

#tag_title#Answer#tag_content# When the mass is released, the angular acceleration of the wheel is given by the formula: \(α = \frac{\tau}{I} = \frac{rF}{\frac{1}{2}MR^2}\). First, we calculate the force exerted by the suspended mass: \(F = mg = (2.00 \mathrm{~kg})(9.81 \mathrm{~m/s^2}) = 19.62 \mathrm{~N}\). Next, we calculate the moment of inertia of the wheel: \(I = \frac{1}{2}(40.0 \mathrm{~kg})(0.3 \mathrm{~m})^2 = 1.8 \mathrm{~kg \cdot m^2}\). Now, we substitute the values of torque and moment of inertia into the formula for angular acceleration: \(α = \frac{(0.3 \mathrm{~m})(19.62 \mathrm{~N})}{1.8 \mathrm{~kg \cdot m^2}} = 3.27 \mathrm{~rad/s^2}\). Therefore, the angular acceleration of the wheel when the mass is released is \(3.27 \mathrm{~rad/s^2}\).