Analyzing torque of an automobile engine

[I_Try_Math]

Homework Statement

An automobile engine can produce 200 Nm 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.

Relevant Equations

$\sum \tau = I\alpha$

My strategy for this problem is to use the equation, $\sum \tau = I\alpha$ to find $\alpha$.
$I_{total} = I_{shaft} + I_{axle} + 2(I_{wheel})$
$I_{wheel} = I_{disk (hub)} + I_{ring (wall)} + I_{ring (treads)}$

Is this the correct way to calculate the moment of inertia?

 

[Lnewqban]

I would calculate the total resistance to accelerated rotation that the 95% of the engine torque will face.

Just be careful with the units.

Please, see:
http://hyperphysics.phy-astr.gsu.edu/hbase/mi.html

 

[I_Try_Math]

I would calculate the total resistance to accelerated rotation that the 95% of the engine torque will face.

Just be careful with the units.

Please, see:
http://hyperphysics.phy-astr.gsu.edu/hbase/mi.html

Does any of my work stand out as being incorrect here?
$I_{shaft} = \frac{1}{2}m_{shaft}r_{shaft}^2$
$= 0.015 kg \cdot m^2$

$I_{axle} = \frac{1}{2}m_{axle}r_{axle}^2$
$= 0.003 kg \cdot m^2$

$I_{wheel} = I_{disk (hub)} + I_{ring (wall)} + I_{ring (treads)}$
$= \frac{1}{2}(m_{disk}r_{disk}^2 + m_{wall}(r_{wall,i}^2 + r_{wall,o}^2) + m_{ring}r_{ring}^2)$
$= 0.92 kg \cdot m^2$

$I_{total} = I_{shaft} + I_{axle} + 2(I_{wheel})$
$I_{total} = 1.86 kg \cdot m^2$

$\sum \tau = I\alpha$
$200 N \cdot m = (1.86 kg \cdot m^2)\alpha$
$\alpha = 102.2 \frac{rad}{s^2}$