Stopping Scooters: Comparing KE & AM

[IsakVern]

TL;DR

When they have equal kinetic energy but not equal angular momentum?

Assume that a kick-scooter rolls on a smooth surface without slipping, and that - for simplicity - all the mass of the scooter's two wheels are distributed like a loop/ring, i.e. around the edges of the wheels with no mass in the centre of the wheels. The wheels have radius R and the scooter is traveling with speed v.

Will another scooter traveling at the exact same speed, with wheels of the exact same mass, but with radii 2R be harder to stop? Because they will have equal kinetic energy but not equal angular momentum.Kinetic Energy:

Scooter 1: KE = 1/2mv2 + 1/2Iw2 = 1/2mv2 + 1/2*mR²*(v/R)² = mv2

Scooter 2: KE = 1/2mv2 + 1/2Iw2 = 1/2mv2 + 1/2*m*(2R)²*(v/2R)² = mv2
Angular momentum:

Scooter 1: L = Iw = mR²*(v/R) = mRv

Scooter 2: L = Iw = m*(2R)²*(v/2R) = m*4R²*v/(2R) = 2mRv

 

[A.T.]

Will another scooter traveling at the exact same speed, with wheels of the exact same mass, but with radii 2R be harder to stop? Because they will have equal kinetic energy but not equal angular momentum.

- Same energy means you have to do the same force * distance.
- Same linear momentum means you have to do the same force * time
- More angular momentum means you have to apply more torque * time, which will happen because of the longer lever arm of the force.

 

[jack action]

Will another scooter traveling at the exact same speed, with wheels of the exact same mass, but with radii 2R be harder to stop?

'Harder' usually refers to force or power. Since power is force times velocity and the velocity is the same in both cases, then we only need to look at the force:

Scooter 1: $F = ma + \frac{I\alpha}{R} = ma + \frac{mR^2 \frac{a}{R}}{R} = 2ma$

Scooter 2: $F = ma + \frac{I\alpha}{2R} = ma + \frac{m(2R)^2 \frac{a}{2R}}{2R} = 2ma$

 

[IsakVern]

'Harder' usually refers to force or power. Since power is force times velocity and the velocity is the same in both cases, then we only need to look at the force:

Scooter 1: $F = ma + \frac{I\alpha}{R} = ma + \frac{mR^2 \frac{a}{R}}{R} = 2ma$

Scooter 2: $F = ma + \frac{I\alpha}{2R} = ma + \frac{m(2R)^2 \frac{a}{2R}}{2R} = 2ma$

So if the scooter is coming towards a person standing still, the person must apply the same force on either scooter in order to stop it, correct?

So does the difference in angular momentum only affect the scooter's ability to stay upright?

 

[A.T.]

So does the difference in angular momentum only affect the scooter's ability to stay upright?

Self stability is it's own can of worms. You can search for old threads on this or checkout this video:

 

[jack action]

Momentum represents the quantity of motion. The greater the mass, the greater the motion. The greater the velocity, the greater the motion. All we know - by observation - is that in any inertial frame it is a conserved quantity.

To change the quantity of motion, you need a force (Newton's second law).

So if the scooter is coming towards a person standing still, the person must apply the same force on either scooter in order to stop it, correct?

Yes, because no matter the radius of your wheel, its radial velocity (not angular velocity) is always the same. And since the mass is the same, then the momentum (quantity of motion) is the same.

If you imagine your wheel as a thin torus, then if you "unbend" it, it will become a cylinder that travels linearly with a velocity $v$. The wheel radius will just affect the length of that cylinder, but not the velocity, not the mass (since it is fixed by definition).

 

[Lnewqban]

Referring only to the practicalities associated to the title of this thread (which does not include the same-mass-wheel condition):
Getting trapped in pavement cracks and pot holes is a big and dangerous problem for scooters with very small wheels.
In that sense, wheels of greater radius are better, but it is not easy to produce bigger wheels while keeping the same mass.

On the other hand, (more applicable to motorcycles than to kick-scooters) wheels with greater angular inertia are harder to accelerate and stop and make steering a little harder.
Font wheels of motocross, enduro and trial motorcycles are made with the biggest practical diameter in order to reduce sinking in sand and to also increase capability to go over rocks and fallen trees.

 

[jack action]

Angular momentum:

Scooter 1: L = Iw = mR²*(v/R) = mRv

Scooter 2: L = Iw = m*(2R)²*(v/2R) = m*4R²*v/(2R) = 2mRv

I figured out what's wrong with this: you forgot the linear component of the momentum:

Scooter 1: L = mv + Iw/R = mv + mR²*(v/R)/R = 2mv

Scooter 2: L = mv + Iw/(2R) = mv + m*(2R)²*(v/2R)/(2R) = mv + m*4R²*v/(2R)² = 2mv

The wheel turns, but it also moves forward.