Rotational inertia of truck and trailer

  • Thread starter CH_GR
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  • #1
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Main Question or Discussion Point

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.
 

Answers and Replies

  • #2
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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.
 
  • #3
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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.
 
  • #4
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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?
 
  • #5
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The wheels weigh 200 lbs for steel rims and 170 lbs for aluminum rims.

Thanks for the info.
 

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