# Physics of decceleration of rolling vehicle

1. Mar 13, 2015

### Wesski

I've been arguing with my physics teacher about why the deacceleration of a person on a skateboard or bicycle that is gliding along the ground (without pedaling or pushing) is dependent on the mass of the person riding it (The higher mass person goes further). I say that it is solely due to air-resistance having the same effect regardless of the mass (in this situation it is a person with 100lbs of lead in a backpack vs the same person without, so no greater area). He says that even in a vacuum, the heavier person riding the skateboard/bike would glide further.
Who is correct and why?

2. Mar 13, 2015

### Stephen Tashi

Are we assuming that the person with the lead and the person without the lead reach the same speed before they quit pushing and begin to glide?

3. Mar 13, 2015

### A.T.

What is his argument/derivation? Assuming that the rolling resistance is proportional to normal force, mass would not affect distance in vacuum.

4. Mar 13, 2015

### Wesski

He says that it isn't really rolling resistance because the skate board has bearings which involve a lot more forces than just a ball rolling. The way I see it the resistance is still proportional to mass so the masses would still cancel out w/o air resistance yes?

5. Mar 13, 2015

### nasu

You need to specify the conditions before you can think about an answer.
As already asked, what is the initial state? Do they have same initial speed or is something else?
How are the resistance forces supposed to depend on the mass of the system?
You have rolling resistance from the ground but also some friction in the bearing which may depend on weight.
A problem insufficiently defined can lead to never-ending "arguments".

6. Mar 14, 2015

### Wesski

The situation is the object starts on and rolls down a ramp about 1m high before gliding to a stop. Sorry for not specifying. I say that (ignoring air resistance) they would travel the same distance because:

They would have the same speed at the end of the ramp because (θ here is angle above horizontal that the ramp is):
ΣF = ma
(cosθ)(m)(g) - Ff = ma
(cosθ)(m)(g) - Crr(N) = ma
(cosθ)(m)(g) - Crr(m)(g) = ma
(cosθ)(g) - Crr(g) = a

And they would have the same acceleration at the bottom of the ramp because:
ΣF = ma
Ff = ma
Crr(N) = ma
Crr(m)(g) = ma
Crr(g) = a

The only way in my mind (barring some mistake here) that one would have a greater or lesser acceleration is if there is another force due to the friction within the bearing, which is either:
a) not proportional to the weight at all
b) is proportional to the weight, but to a different degree (i.e., √N, or N2).
Otherwise it would cancel, correct?

7. Mar 14, 2015

### billy_joule

That still doesn't really specify the conditions. If you and your teacher treat the wheels and bearings differently then you'll get different results.

IME deformation of the wheels is a significant source of energy loss. After skating down a hill all wheels are hot. An unloaded (runaway) skateboard can roll further than a loaded skateboard - it depends on the wheel compound rating and the axle bolt torque (ie if the inner race is held solid or free to rotate about the axle). Anecdotal so take with a grain of salt.

8. Mar 14, 2015

### dean barry

Assuming both are the same physical size but different masses
1)
The rolling resistance force ( deemed to be constant regardless of speed ) is proportional to the mass.
( greater mass = greater rolling resistance force )

2) the air drag force is the same for both

Consider both forces negative as they oppose motion

So:
Since deceleration = force / mass
The heavier mass will have the greater deceleration.

9. Mar 14, 2015

### nasu

Yes, with this kind of friction they will travel the same distance.
However your expression for acceleration on the incline is wrong.
what is the normal force there?

10. Mar 14, 2015

### A.T.

If you assume that resistance is proportional to mass, then yes. But that is just an approximation.

11. Mar 14, 2015

### A.T.

Better still: Do the experiment with your teacher.

12. Mar 14, 2015

### Wesski

Oh whoops. The normal would be cosθ*mg and the downward force would be sinθ*mg. My mistake. However mass still cancels out.
Is it possible in some cases for additional mass to cause the coefficient of rolling resistance to be lower or higher?

13. Mar 15, 2015

### A.T.

Yes. Constant coefficient of rolling resistance is just an approximation.

14. Mar 21, 2015

### dean barry

i input this onto excel with two arbitrary masses and the deceleration rate is the same regardless of the mass.

The decelerating force = m * g * cosine (incline angle) * Crr
The accelerating force = m * g * sine (incline angle)

Juggle the Crr value until you get a net deceleration, then change the mass, the deceleration rate remains the same.

15. Mar 21, 2015

### A.T.

You needed excel to see this?

Yes, assuming Crr is independent of the mass, which is not exactly true in nature.