Rotational inertia of truck and trailer

Click For Summary

Discussion Overview

The discussion centers on calculating the energy required to accelerate a truck and trailer from rest to 60 mph, considering the mass of the vehicle and the rotational inertia of the wheels and rims. The scope includes theoretical calculations related to dynamics and energy, with a focus on the contributions of the wheels to the overall kinetic energy.

Discussion Character

  • Technical explanation
  • Mathematical reasoning

Main Points Raised

  • One participant inquires whether to treat the rims and tires as a cylindrical shell or a solid cylinder for energy calculations.
  • Another participant provides the moment of inertia formulas for both cylindrical shell and solid cylinder, suggesting that the wheels can be approximated as 0.8mR² for calculations.
  • A question is raised about whether this approximation applies universally to all passenger car and truck tires.
  • A participant confirms that the tire and rim together can be approximated as I = 0.8mR² and notes that the wheels constitute a small percentage of the total mass of the truck and trailer.
  • Further calculations are presented for the kinetic energy of the wheels, leading to a total kinetic energy equation that includes contributions from both the vehicle and the wheels.
  • The weight of the wheels is clarified, with specific weights provided for steel and aluminum rims.

Areas of Agreement / Disagreement

Participants generally agree on the approximation of the wheels' moment of inertia and the method for calculating total kinetic energy, but there is uncertainty regarding the applicability of this approximation to all types of tires.

Contextual Notes

The discussion does not resolve the assumptions regarding the uniformity of tire construction and its impact on the calculations. There are also unresolved questions about the specific weights of the wheels and how they factor into the overall energy computation.

CH_GR
Messages
3
Reaction score
0
I need to compute the energy to get a truck and trailer from 0 to 60 mph from rest taking into account the mass of truck and trailer and inertia of wheels and rims.

Do I treat the rims and tires as a cylindrical shell or solid cylinder? Once I find the energy of the tires and rims do I just add that to the 1/2MV^2 for truck and trailer?

thanks.
 
Physics news on Phys.org
The moment of inertia of a cylindrical shell is mR2, while the solid cylinder is (1/2)mR2. The wheels are about 0.8mR2. For an automobile, the rolling tires add about 4% to the dynamic mass e.g., kinetic energy KE = (1/2)(1.04) mv2, where m is the static mass.
 
By wheel you mean tire and rim? Then I could use 0.8MR^2 to approximate the inertia instead of treating it separately as cylindrical shell and solid shell?

Does this approximation apply to all passenger car and truck tires?

thanks.
 
Yes, the tire and rim together are about I = 0.8 mR2. The 18 tires plus rims (=wheels) are only a small percentage (<5%) of total truck & trailer mass.
So the kinetic energy of a wheel is
KE = (1/2) I w2 = (1/2)(0.8)m (Rw)2 = (2/5) mv2 (rotational energy only)
For 18 wheels it is
KE = (36/5)mv2
This gets added to the total vehicle kinetic energy:
kE tot = (1/2)Mv2 + (36/5) mv2
How much do the wheels weigh (mg)? 200 pounds?
 
The wheels weigh 200 lbs for steel rims and 170 lbs for aluminum rims.

Thanks for the info.
 

Similar threads

  • · Replies 18 ·
Replies
18
Views
3K
  • · Replies 10 ·
Replies
10
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 49 ·
2
Replies
49
Views
6K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 60 ·
3
Replies
60
Views
7K
  • · Replies 11 ·
Replies
11
Views
2K
Replies
48
Views
14K