Torque required at the wheels to accelerate a vehicle at a certain rate?

[Jules575]

TL;DR Summary

Struggling to reconcile Newton's second law for linear and rotational acceleration. How are these related for wheels accelerating (linear) a vehicle?

I'm struggling to understand something basic here. If I have a just a wheel, with mass 10kg, and radius 0.25m, and I specify that the CG is accelerating linearly at 1ms^-2, how do I calculate the force needed to do this? Using F = ma gives 10N, but using this value for torque calculation on the wheel gives
T = 10×0.25 = 2.5Nm
α = T/I where I = 0.5×10kg×(0.25m)² = 0.3125 kgm²
α = 2.5/0.3125 = 8 rads^-2
and lastly, the linear acceleration of the CG of the wheel is given by
a = α×r = 8×0.25 = 2ms^-2
This is double what I needed, why does using the force from Newton's second law for linear motion not agree with the law for rotational motion?
Any help greatly appreciated!

 

[hutchphd]

If you push the wheel on a slippery surface (no friction) it will not rotate and it will require 10 N force to produce the 1ms^-2.
If there is enough friction to make the wheel rotate without slipping how big is that frictional force? How hard will you now need to push to get the 1ms^-2?

 

[Jules575]

If you push the wheel on a slippery surface (no friction) it will not rotate and it will require 10 N force to produce the 1ms^-2.
If there is enough friction to make the wheel rotate without slipping how big is that frictional force? How hard will you now need to push to get the 1ms^-2?

Intuitively, I am inclined to say that you don't need to exert any additional force just because of the presence of static friction at the instant centre, but honestly I'm driving myself up the wall so I could just be way off.

 

[hutchphd]

The ground will exert a force (or torque if you prefer) on the wheel in order to make it rotate. What is the direction of that force?? How big must it be to get the desired angular acceleration. Numbers please and maybe a free body diagram...I think you already worked them out.

 

[jbriggs444]

If you pull the rug out from under a person's feet, do the feet move at the same or different speed from the person?

Edit: Possibly I have failed to understand the scenario.

Are we talking about accelerating a wheel by pulling on the road underneath it? Or are we talking about accelerating the wheel by applying an external torque and letting the stationary road accelerate it? In the former case, the roll rate will not match the vehicle movement rate and, in fact, they will be in opposite directions. In the latter case, the torque from the road and the external torque from the motor will not match and again, they will be in opposite directions.

 

[DaveE]

As others have said, I'll repeat in a slightly different way.
It takes force to make the vehicle move (let's say in the frictionless sense, like a hockey puck). It also takes force to make a wheel spin, even if there is no linear motion (like a bicycle wheel held off the ground). You have to do both to get your vehicle to move. Some of the energy you apply goes into linear motion, some goes into rotary motion. As you already know, the tire touching the ground (without slipping) determines the ratio between the two.

 

[jack action]

$$T = mar + I\alpha$$
$$T = \left(1 + \frac{I\alpha}{mar}\right)mar$$
$$T = \left(1 + \frac{I}{mr^2}\right)mar$$
Or:
$$F = \left(1 + \frac{I}{mr^2}\right)ma$$
And if $I = \frac{1}{2}mr^2$, then:
$$F = \left(1 + \frac{\frac{1}{2}\rlap{/\ /}mr^2}{\rlap{/\ /}mr^2}\right)ma$$
$$F = \left(1 + \frac{1}{2}\right)ma$$
$$F = 1.5ma$$

So 2/3 ("1") of the force is for linearly accelerating the wheel and 1/3 ("½") is for the rotational acceleration.

In your example, it means you need 15 N to accelerate the wheel: 10 N (= 10 kg X 1 m/s²) + 5 N (= 0.3125 kg.m² X 1 m/s² / (0.25 m)²)