rotational vs linear acceleration

[David]

First let me point out the attached cartoon. I have not studied math or physics since 1982. Someone PM'd me this question below. I have replied in an open space as there may be people here with fresher physics skills then me.

"I was wondering if you might know the answer to a question I can't find an answer for. Is there a difference in energy required to rotate a 50kg flywheel at the same acceleration rate and speed as it would take to pull a 50kg weight at the same rate and speed? Friction being equal between the two."


For acceleration in a straight line, F=ma. Force = mass X acceleration. I actually remember that. 50KG X 9.8 ms^2 (1 G of acceleration) = 490.5 N

For rotation, T = IA T = torque = force X radius. I = rotational inertia = 0.5mr^2, where r is the outside radius of a disk. A = rotational acceleration in revolutions per second squared. Torque = Fr, ie force X radius. T = Fr = 0.5mr^2

If I take the same 490.5 N of force and 50 KG of mass, 490.5r = 0.5X50r^2. Assume 1 m radius and I get 19.62 rev/s^2. This is of course applies to oranges. I think he wanted to know what G level the flywheel would have.

For me there is no clean answer to the question. Can anyone else do better?

If it matters, the equation for acceleration of a flywheel's outer surface at a constant rpm is g=4.02568rf, where r = radius and f = revs/second

[CDave]

Thanks David. What I am curious about is, would the power requirement be the same.

What got me wondering is the tires on my truck weigh 12lbs. more, per tire, than the O.E. size of the same brand-model. I was curious if the amount of power to turn those heavier tires would be the same as having 48lbs. extra weight in the bed of my truck. If the amount of friction were equal between the two loads.

Say your engine has 300hp. What would use more of the available power, friction being equal. The tires being 48lbs. heavier or carrying a 48lb. load in the bed.
Would power requirement for acceleration and traveling at a steady speed be the same amount, between the two, in each situation.

[CDave]

This?
https://www.engineeringtoolbox.com/m...las-d_941.html

[David]

I might be able to do a rough calculation, assuming that traction is not an issue. What is the diameter of the new and old tires? What is the weight of the new and old/wheel tire assemblies? What does the truck weigh?

[David - WI]

Would power requirement for acceleration and traveling at a steady speed be the same amount, between the two, in each situation.

The wheels & tires don't use any power when the vehicle is traveling at a steady speed (if you''re ignoring "friction" / rolling / wind resistance of the tires).

Accelerating... you have to add tbe power it takes to accelerate the extra 48 pounds up to whatever mph you want to go, plus the power it takes to spin the the extra weight up to whatever rpm they will be turning at that mph. But, in theory you get that power back when you slow down without using the brakes... you will coast that much further/faster with the heavier tires.

[CDave]

Thanks everyone, I was just curious if there was a diff. in the two different types of weight.
I understand my new tires add friction being they are wider and taller. The taller the tire the larger the contact patch, plus the added width make a larger contact patch.
I also understand wheels/tires act like flywheels and like David - WI said you get more stored energy for coasting with the heavier set.

I like physics but I, unlike my dad, am terrible at math. I like to say I have math dyslexia.

[David]

As it turns out, this is not the only site I frequent. I posed the same question on a car web site. A fellow engineer there said that the multiplier is about 1.75. Adding weight to your wheel/tire assembly has a bigger effect on acceleration then adding weight in the bed as you have to get the additional mass rotating as well as translating forward.

So 40 lb of extra wheel tire weight has about the same effect on acceleration as 70 lb in the bed. Diddly squat in a 5000-6000 lb pick up.

I do wish I could do the math or see the math, but not yet.

[CDave]

Thanks David, I thought it would cost more hp accelerating a flywheel than just hauling the same amount of weight, but didn't know for sure. Thanks for the multiplier that's a fun bit of info to have. My truck is approx. 4500lbs.

[David]

It's a rainy day

Assume the truck weighs 4500 lb, the base wheel and tire is 30 kg (66 lb), the upgrade wheel and tire weighs 35 kg
Pick a tire size of 245/60-18 (actual size for a Honda Ridgeline, which weighs about 4500 lb)
Assume 100 kph

translational KE = 0.5mV^2 = 771605 Joules (I refuse to do physics in US units)

rotational KE per disk = 0.5Iw^2 I = moment of inertial = 0.5mR^2 w = rotational speed
KE per disk = 587 J
KE for 4 = 2350J

total KE = 789844 J
% of total KE from rotating tires = 0.3%

Increase tire mass to 35 kg and the % of total KE from rotating tires is only 0.35%

I think the multiplier I was given on the other web site is too high.

[CDave]

Thanks again. I went from a LT245/75/16 to LT305/70/16. Which is approx. 30-10-16 to 33-12.5-16 and 48lbs. to 62lbs.
So it's 14lbs. diff per tire not the 12lbs. I thought it was.

[David - WI]

Bigger diameter tire hurts acceleration... just like a higher pitch prop. Plus makes speedo & odometer read low so you have to compensate when you calculate fuel mileage (or correct with programmer like HyperTech that also lets you re-adjust the shift points).

[Stoker boy]

For me there is no clean answer to the question.

Take the vehicle or vessel in question, try different flywheels run it through a timed track until you get the results you desire.