Rotating Mass when keeping a minimum weight

In summary, the rotational kinetic energy of a weight distributed on a rotating wheel is greater than the rotational kinetic energy of a weight distributed on a non-rotating wheel, but the increase in rotational kinetic energy is negligible depending on the weight distribution of the wheel.
  • #1
huesv
4
0
I've looked at the equations for torque, inertia, and angular acceleration, but I still can't figure this out. I was hoping someone could push me in the right direction.

This is the question I'm trying to solve:
In a situation where an engine propels a required amount of weight, is a reduction in rotating mass really beneficial (in non-constant RPM situations) because any weight savings in rotating mass is added back as static weight.

For instance, let's say a 1000 kg vehicle had an engine driving a single disk (solid, uniform density) rear wheel. If a 10 kg weight savings is made to the rear wheel, that 10kg must be added back to the vehicle as static weight so that the vehicle weighs 1000 kg total.

I believe that the engine will be able to angularly accelerate the lighter rear wheel faster, but since it's propelling more static weight, will the net advantage be positive, negative, or negligible?

Thanks in advance!
 
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  • #2
Resistance to angular acceleration is moment of inertia, which doesn't just depend on the weight of the disk, it depends on the distribution of the weight of the disk. So the answer is that it depends on the geometry of the disk.
 
  • #3
russ_watters said:
Resistance to angular acceleration is moment of inertia, which doesn't just depend on the weight of the disk, it depends on the distribution of the weight of the disk. So the answer is that it depends on the geometry of the disk.

Lets say the disk (rear wheel) is solid and of uniform density. Let's also assume that the weight savings of the disk is made by decreasing the density of the disk. So that the dimensions are constant but it weighs less, e.g. Mdisk > (Mlightdisk = Md/4).
 
  • #4
A wheel of a vehicle both rotates and moves with translation (center of gravity moves with the speed of the vehicle) . It's kinetic energy is a sum of translational and rotational kinetic energy (you can imagine accelerating the wheel in two phases: first you accelerate it's axis, then rotate it), so it is obviously greater than it would be without rotation.
 
  • #5
Lojzek said:
A wheel of a vehicle both rotates and moves with translation (center of gravity moves with the speed of the vehicle) . It's kinetic energy is a sum of translational and rotational kinetic energy (you can imagine accelerating the wheel in two phases: first you accelerate it's axis, then rotate it), so it is obviously greater than it would be without rotation.

I think you're saying that it takes more energy to move a point on the wheel because the engine must accelerate it angularly and translationally. So, regardless of the weight distribution of the disk, it will always take more energy to move rotating weight than non-rotating weight (in this scenario). That's what I thought, but can someone give me a set of equations so I can figure out if it's negligible in my case?
 
  • #6
If the speed of the vehicle is v, then most points of the wheel rotate aroud wheel's axis with slower speed than v. So if you transfer a mass m from wheel to fixed mass of the vehicle, the decrease of kinetic energy would be less than m*v^2/2.
Exact value depends on weight distribution of the wheel:

Formula for rotational kinetic energy is: E=J*omega^2/2, where

J(moment of inertia)=Integral (r^2*dm)

In case of a ring (all mass at maximum radious) J would be m*R^2, which gives rotational energy equal to m*v^2/2 (v=omega*R). In case of a uniform cylinder J is m*R^2/2, which gives half less rotational energy.
 
  • #7
Ok, excellent. So... would this be right?

KEtot = KEr + KEt
(Mw*Vw^2)/2 = (Mr*R^2*(2*pi*ƒ)^2)/2 + (Mo*Vo^2)/2

where
Mw = mass + extra weight (fixed)
Mo = mass (without extra weight fixed)
Mr = mass of extra weight (which will rotate)
ƒ = revs per second

and assuming the extra weight is a ring.

If I know the velocity at ƒ, Vo, and all the masses and R, then I can see what my speed will be, Vw, if I moved the rotating weight to a fixed position; right?
 
  • #8
True. You can also eliminate one variable with the equation v0=2*pi*R*f (if we are talking about a wheel which must touch the ground).
 

Related to Rotating Mass when keeping a minimum weight

1. What is rotating mass?

Rotating mass refers to the weight of an object or system that is in motion, typically rotating around an axis. This can include objects such as wheels, gears, or propellers.

2. Why is it important to keep a minimum weight when dealing with rotating mass?

Keeping a minimum weight for rotating mass is important because it affects the overall performance and stability of the system. A heavier rotating mass can require more energy to move and can also cause increased wear and tear on the system.

3. How does rotating mass affect acceleration?

The greater the rotating mass, the more inertia it has. This means that a heavier rotating mass will be more resistant to changes in speed and direction, resulting in slower acceleration.

4. Can rotating mass impact fuel efficiency?

Yes, rotating mass can have an impact on fuel efficiency. A lighter rotating mass will require less energy to rotate, resulting in less fuel consumption. This is especially important in vehicles and machinery that require a lot of power to move.

5. How can we reduce the weight of rotating mass without sacrificing performance?

There are a few ways to reduce the weight of rotating mass without compromising performance. These include using lighter materials, optimizing the design to reduce unnecessary weight, and finding ways to redistribute weight to achieve better balance and stability.

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