Analyzing torque of an automobile engine

AI Thread Summary
The discussion focuses on calculating the moment of inertia for an automobile engine using the equation ∑τ = Iα. The total moment of inertia is derived from the individual components: shaft, axle, and wheels, with specific calculations provided for each. Participants emphasize the importance of unit consistency and precision in calculations, recommending to maintain at least three decimal places for accuracy. Corrections are made regarding the proper identification of components, particularly the tread of the tire as a hoop. Ultimately, the original poster successfully arrives at the correct answer after addressing the feedback.
I_Try_Math
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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?
 
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Lnewqban said:
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} ##
 
I_Try_Math said:
Does any of my work stand out as being incorrect here?
Yes.
I_Try_Math said:
## I_{wheel} = I_{disk (hub)} + I_{ring (wall)} + I_{hoop (treads)} ##
"The tread of each tire acts like a ... hoop of radius 0.330-m."

Wheel moments of inertia.jpg
 
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In case you are puzzled by @Lnewqban's post, check the scope of your parentheses…
I_Try_Math said:
## = \frac{1}{2}(… + m_{ring}r_{ring}^2) ##
Also, I would keep each term to at least three decimal places until they are summed.
 
Lnewqban said:
Yes.

"The tread of each tire acts like a ... hoop of radius 0.330-m."

View attachment 349070
You're correct and thank you. I was finally able to get the right answer.
 
haruspex said:
In case you are puzzled by @Lnewqban's post, check the scope of your parentheses…

Also, I would keep each term to at least three decimal places until they are summed.
Right, thank you for the hints. I was finally able to get the correct answer.
 
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