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

• Jules575
In summary: N.In summary, when calculating the force needed to accelerate a wheel with a given mass and radius at a certain rate, it is important to consider both the linear and rotational aspects of the motion. Using the formula F = ma for linear motion may not give the same result as using the formula T = Iα for rotational motion because the force is divided between the two types of motion.
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!

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?

hutchphd said:
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.

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.

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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.

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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.

hutchphd
$$T = mar + I\alpha$$
$$T = \left(1 + \frac{I\alpha}{mar}\right)mar$$
$$T = \left(1 + \frac{I}{mr^2}\right)mar$$
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)

## 1. What is torque and how does it affect a vehicle's acceleration?

Torque is a measure of the force that causes an object to rotate. In the context of a vehicle, it is the force that the engine generates to turn the wheels. The greater the torque, the faster the vehicle can accelerate.

## 2. How is torque calculated for a vehicle?

Torque is calculated by multiplying the force applied to the wheels by the distance from the center of the wheel to the point where the force is applied. This distance is known as the lever arm. The unit of torque is Newton-meters (Nm).

## 3. What factors affect the torque required to accelerate a vehicle at a certain rate?

The main factors that affect the torque required for acceleration are the weight of the vehicle, the gear ratio of the transmission, and the rolling resistance of the tires. The steeper the incline or the heavier the load, the more torque is needed to maintain a certain rate of acceleration.

## 4. How does torque differ from horsepower?

Torque and horsepower are often used interchangeably, but they are not the same thing. Torque is a measure of rotational force, while horsepower is a measure of power or work over time. In simple terms, torque determines how quickly a vehicle can accelerate, while horsepower determines how fast it can maintain that acceleration.

## 5. Can torque be increased to improve a vehicle's acceleration?

Yes, torque can be increased through modifications to the engine, such as adding a turbocharger or supercharger. However, it is important to note that increasing torque also puts more strain on the engine and can lead to potential mechanical issues if not done properly.

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