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# An automobile engine can produce 200 N · m of torque. Calculate the angular acceleration produced if 95.0% of this torque is applied to the drive shaft, axle, and rear wheels of a car, given the following information. The car is suspended so that the wheels can turn freely. Each wheel acts like a 15.0-kg disk that has a 0.180-m radius. The walls of each tire act like a 2.00-kg annular ring that has inside radius of 0.180 m and outside radius of 0.320 m. The tread of each tire acts like a 10.0-kg hoop of radius 0.330 m. The 14.0-kg axle acts like a rod that has a 2.00-cm radius. The 30.0-kg drive shaft acts like a rod that has a 3.20-cm radius.

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First, we need to calculate the total moment of inertia of each type of component. In the case of the wheels, the torque is applied to two of them and so, the total moment of inertia of the wheels $I_{\text{wheels}}$ will be equal to two times the moment of inertia of one wheel. Similarly, we have two sets of wheel walls, therefore $I_{\text{walls}}$ will be equal to two times the moment of inertia of one set of walls. In the same way, there are two treads (one per wheel), so we proceed in the same way to calculate $I_{\text{treads}}$. Finally, there is only one axle and one shaft so we don't have to multiply $I_{\text{axle}}$ and $I_{\text{shaft}}$ by any factor.

\begin{align*} I_{\text{wheels}}&=2\left(\dfrac{M_{\text{wheel}}{R_{\text{wheel}}}^2}{2}\right) \\ &=M_{\text{wheel}}{R_{\text{wheel}}}^2 \\ &=\left(15.0\right)\left(0.180\right)^2 \\ &=0.486 \text{ kg}\cdot {\text{m}}^2 \\ I_{\text{walls}}&=2\left[\dfrac{M_{\text{wall}}}{2}\left({R_2}^2+{R_1}^2\right)\right] \\ &=\left(2.00\right)\left[\left(0.320\right)^2 +\left(0.180\right)^2\right] \\ &=0.270 \text{ kg}\cdot \text{m}^2 \\ I_{\text{treads}}&=2 M_{\text{tread}}{R_{\text{tread}}}^2 \\ &=2\left(10.0\right)\left(0.330\right)^2 \\ &=2.178 \text{ kg} \cdot \text{m}^2 \\ I_{\text{axle}}&=\dfrac{M_{\text{axle}}{R_{\text{axle}}}^2}{2} \\ &=\dfrac{\left(14.0\right)\left(0.0200\right)^2}{2} \\ &=0.0028 \text{ kg}\cdot \text{m}^2 \\ I_{\text{shaft}}&=\dfrac{M_{\text{shaft}}{R_{\text{shaft}}}^2}{2} \\ &=\dfrac{\left(30.0\right)\left(0.0320\right)^2}{2} \\ &=0.0154 \text{ kg}\cdot \text{m}^2 \\ \end{align*}

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