Analyzing torque of an automobile engine

  • #1
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} ##
 
  • #4
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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  • #5
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.
 
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  • #6
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.
 
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  • #7
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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