# Torque on accelerating tire

So I have a problem understanding the torques on a tire as it accelerates from rest. Suppose I have a tire with mass $$m = 10kg$$, radius $$r = 0.5m$$ and I apply torque with magnitude $$\tau = 30Nm$$ to its axis. The force with which the tire pushes on the ground should be equal to $$\tau / r$$, so 60N. The maximum friction force $$F_{f(max)} = \mu m g = 78.4N$$, where $$\mu = 0.8$$ and $$g = 9.8m/s^2$$ so the force from the tire doesn't exceed the maximum friction force. And that means it won't slip. According to Newton's third law, the force produced by the tire is counteracted by an equal in magnitude and opposite in direction friction force $$F_f = -F = 60N$$. That friction creates torque about the axis of rotation which is in opposite direction to the torque I started with and has magnitude $$\tau_f = F_f r = 30Nm$$ so they should completely cancel out, leaving 0 net torque on the tire and that means it won't rotate. I must be wrong somewhere because the tire will obviously accelerate when a torque is applied to it. I've read somewhere that some of the torque goes for the angular acceleration of the tire and not all of it produces the force at the bottom of the tire. Is that in any way correct?

Any help would be appreciated!

• N3L - force on object A due to object B is equal in magnitude and opposite in direction to force on object B due to object A, ie the forces are on different objects. Commented Jan 15, 2020 at 9:36

## 4 Answers

I apply torque with magnitude 𝜏=30Nm to its axis.

Sure.

The force with which the tire pushes on the ground should be equal to 𝜏/𝑟, so 60N.

If the wheel were massless, that would be true. 100% of the torque from the engine would be "delivered" to the ground. But the wheel is not massless. Some of the torque is instead going into accelerating the wheel.

Using the sum of torques formula:

$$\tau_{net} = I\alpha$$

If the wheels rotation is accelerating, then there must be a non-zero net torque. Therefore the torque from the axle and the torque from the ground cannot be equal.

There is a force pair where the force from the tire on the ground is exactly equal to the force from the ground on the tire. But when examined as a torque, these will be less than the torque from the engine.

One way to think about this is the linear example of pushing two blocks forward. You push on A, and A pushes on B. If A is massless, B is pushed with an equal magnitude force.

But the more mass A has, the smaller the force A pushes on B. In your car scenario, the larger the moment of inertia of the wheel, the smaller the torque that the ground creates.

And that would mean that if the wheel were massless, then the torque would be zero and it would not roll, right?

No. A massless wheel can roll or accelerate. It just cannot have any net torque applied. Any torque placed on it must be countered by an equal torque elsewhere. It's an ideal limit.

But is there a way to precisely calculate the force at the bottom from engine torque apart from the formula for net torque?

You have to solve simultaneous equations for the system. You know the moment of inertia for the wheel, and you know the mass of the car.

The torque by the ground on the wheel will be an amount between zero and the engine torque, and that allows the wheel acceleration to match the vehicle acceleration.

If the torque is too high, the vehicle moves faster than the wheel turns. If the torque is too low, the wheel spins faster than the vehicle moves.

$$\alpha = \frac{a}{t}$$ $$F_{road} = ma$$ $$\tau_{engine} - (F_{road} \times r) = I\alpha$$

• And that would mean that if the wheel were massless, then the torque would be zero and it would not roll, right?
– John
Commented Jan 16, 2020 at 6:58
• Finally an answer that makes sense! Thank you very much! But is there a way to precisely calculate the force at the bottom from engine torque apart from the formula for net torque?
– John
Commented Jan 16, 2020 at 7:01
• Mate any chance you can take a look at this? engineering.stackexchange.com/questions/55295/… You wouldn't believe I've got a Physics degree would you... I was never any good at this mechanics stuff Commented May 19, 2023 at 20:05

Let us consider the particle (or a sum of particles) in contact with the surface. The torque generates force (=60N) on the particle (or sum of particles) in the backward direction or in the direction opposite to the direction in which it intends to move. But the friction acts on the same particle (=60 N) causing the net force on the particle to be 0.

[Please note here that this is not the case of forces on different bodies mentioned in previous answer. This is because the axle producing torque applies a force on the tire, and the friction also applies force on the particle of the tire in contact with it in opposite direction. For example, when pushing a box on a surface with a force less than the max friction, the box will not move as the friction and the force applied by you cancel each other out, both being applied on the box.]

As mentioned above, the particle does have a force = 0N and that is why it acts as an axis of rotation without slipping. The frictional force does not act on the rest of the particles and the torque causes them to rotate about the axis (ie. the particle at rest at that instant) causing it to move forward.

Your logic would have been correct in some cases of linear motion like the case of box mentioned above. In a case like walking, the force by the leg on the floor in the backward direction and the friction acting in the forward direction are applied on different bodies and thus, do not cancel each other out. These two cases should not be confused with one another.

• What happens to the torque? What is it equal to? And how does the friction not act on the whole wheel. What accelerates it then?
– John
Commented Jan 15, 2020 at 16:26
• What do you mean by what happens to the torque? It is equal to 30 N/m. As I said, in linear motion, the friction will act on the whole but in rotation, the part where it acts gets converted to an axis and the torque accelerates the wheel. See it this way- If you ride a bike and pull the front brakes, it would stop the translational (linear) component of the bike. But it would do a wheelie right? That means it is still able to do a rotation even though the brake is applied. The wheel gets converted into an axis. Compare the wheel to the particle and the whole bicycle to the wheel.
– Scar
Commented Jan 15, 2020 at 16:34
• But see that the particle acts as the axis for just an instant. The rotation about that axis causes the point of contact to change to the adjacent particle in the next instance. Thus the car moves as a whole.
– Scar
Commented Jan 15, 2020 at 16:37
• So if I understood you correctly...The torque would be the same 30 Nm, the force that acts on the center of mass would be equal to the friction force which in turn will produce $6 m/s^2$ linear acceleration and because the wheel is not slipping the following is true $\alpha = a / r = 12 rad/s^2$. The moment of inertia of the wheel is $1/2mr^2 = 1.25 kgm^2$ and the torque which is $\tau = \alpha * I = 15 Nm$ and that is half the torque which I started with. What's going on?
– John
Commented Jan 15, 2020 at 16:45
• Why would the force on the center of mass be equal to the frictional force? The friction just converts the particle in contact as a stationary axis for rotation for an instant. The force on the center of mass will be equal to the force produced by the torque. Did you understand the bike analogy?
– Scar
Commented Jan 15, 2020 at 16:49

The only role friction has here is to prevent slipping. It does not influence the linear acceleration. Here is why:

The total KE is equal to the sum of KE of rotation and KE of translational motion. See that both of them are provided by the torque which is generated by the axle. So, the force is provided by the car engine, so to speak.

• And what is "the force provided by the car engine" equal to? Give me concrete numbers for my example. Thanks in advance!
– John
Commented Jan 15, 2020 at 17:09
• 60 N. It has already been covered.
– Scar
Commented Jan 15, 2020 at 17:20
• But that is also the magnitude of the friction force. And the friction force acts in the direction of motion.
– John
Commented Jan 15, 2020 at 17:22
• Your point being?
– Scar
Commented Jan 15, 2020 at 17:28
• You said that the friction force is not the one accelerating the wheel, but the force produced by the wheel itself. But that could not be the case because the force the wheel exerts on the road is in the opposite direction of motion.
– John
Commented Jan 15, 2020 at 17:31

You are making the common mistake of thinking that Newton's third law means that forces should cancel each other resulting in no net force and nothing accelerating.

In order to consider the effect of the equal and opposite forces bodys exert on each other according to Newton's third law, you need to apply Newton's second law to each body individually. Although the magnitude of the forces on the two bodies may be the same, it is the net force on each body individually divided by the mass of each body that determines its acceleration, per Newton's second law.

Now take the example of the wheel. If you draw a free body diagram of the wheel you will see that the only external force acting on the wheel (ignoring other possible dissipative forces) is the static friction force acting forward. That is the net force acting on the wheel and the force responsible for the acceleration of the wheel.

I'm actually more interested about what happens to the torque on the wheel. Is it 0? If not, what is it equal to?

The torque on the wheel is obviously not zero because that would mean the opposing static friction force would also be zero resulting in no acceleration. What happens is the torque on the wheel is transmitted to the ground (the Earth).

A free body diagram of the ground would show the that the clockwise torque of the wheel would be doing work on the earth by causing a counter clockwise torque on the Earth. Technically, that gives the Earth a counter clockwise angular acceleration of

$$α=\frac{τ}{I}$$

Where $$τ$$ is the torque applied to the earth by the wheel, and $$I$$ is the moment of inertia of the Earth. But the moment of inertia of the Earth is so high (about 8 x 10$$^{37} kg.m^2$$) that the angular acceleration is infinitesimal, so it can be ignored.

Hope this helps.

• Comments are not for extended discussion; this conversation has been moved to chat. Commented Jan 15, 2020 at 21:50
The passage primarily discusses the physics of a tire accelerating from rest, focusing on the torques and forces involved, and includes a detailed explanation of the role of friction and the distribution of torque between the tire and the ground.
no_question

the passage does not contain a direct mathematical question that requires a specific numerical answer. instead, it provides a detailed explanation and discussion of the physics involved in the scenario. while there are calculations and numerical values mentioned, they are used to support the explanation rather than to pose a question.
Given a text passage, determine:
    1. If it contains a mathematical question (including word problems or calculations)
    2. If an answer to that mathematical question is provided in the text

First, describe the content structure you found in 1-2 sentences.

Then provide your analysis in the following format:
TYPE: <one of the following values>
- NO_QUESTION
- SINGLE_QUESTION
- SINGLE_QUESTION_WITH_ANSWER
- MULTIPLE_QUESTIONS
- MULTIPLE_QUESTIONS_WITH_ANSWERS
system
# Torque on accelerating tire

So I have a problem understanding the torques on a tire as it accelerates from rest. Suppose I have a tire with mass $$m = 10kg$$, radius $$r = 0.5m$$ and I apply torque with magnitude $$\tau = 30Nm$$ to its axis. The force with which the tire pushes on the ground should be equal to $$\tau / r$$, so 60N. The maximum friction force $$F_{f(max)} = \mu m g = 78.4N$$, where $$\mu = 0.8$$ and $$g = 9.8m/s^2$$ so the force from the tire doesn't exceed the maximum friction force. And that means it won't slip. According to Newton's third law, the force produced by the tire is counteracted by an equal in magnitude and opposite in direction friction force $$F_f = -F = 60N$$. That friction creates torque about the axis of rotation which is in opposite direction to the torque I started with and has magnitude $$\tau_f = F_f r = 30Nm$$ so they should completely cancel out, leaving 0 net torque on the tire and that means it won't rotate. I must be wrong somewhere because the tire will obviously accelerate when a torque is applied to it. I've read somewhere that some of the torque goes for the angular acceleration of the tire and not all of it produces the force at the bottom of the tire. Is that in any way correct?

Any help would be appreciated!

• N3L - force on object A due to object B is equal in magnitude and opposite in direction to force on object B due to object A, ie the forces are on different objects. Commented Jan 15, 2020 at 9:36

## 4 Answers

I apply torque with magnitude 𝜏=30Nm to its axis.

Sure.

The force with which the tire pushes on the ground should be equal to 𝜏/𝑟, so 60N.

If the wheel were massless, that would be true. 100% of the torque from the engine would be "delivered" to the ground. But the wheel is not massless. Some of the torque is instead going into accelerating the wheel.

Using the sum of torques formula:

$$\tau_{net} = I\alpha$$

If the wheels rotation is accelerating, then there must be a non-zero net torque. Therefore the torque from the axle and the torque from the ground cannot be equal.

There is a force pair where the force from the tire on the ground is exactly equal to the force from the ground on the tire. But when examined as a torque, these will be less than the torque from the engine.

One way to think about this is the linear example of pushing two blocks forward. You push on A, and A pushes on B. If A is massless, B is pushed with an equal magnitude force.

But the more mass A has, the smaller the force A pushes on B. In your car scenario, the larger the moment of inertia of the wheel, the smaller the torque that the ground creates.

And that would mean that if the wheel were massless, then the torque would be zero and it would not roll, right?

No. A massless wheel can roll or accelerate. It just cannot have any net torque applied. Any torque placed on it must be countered by an equal torque elsewhere. It's an ideal limit.

But is there a way to precisely calculate the force at the bottom from engine torque apart from the formula for net torque?

You have to solve simultaneous equations for the system. You know the moment of inertia for the wheel, and you know the mass of the car.

The torque by the ground on the wheel will be an amount between zero and the engine torque, and that allows the wheel acceleration to match the vehicle acceleration.

If the torque is too high, the vehicle moves faster than the wheel turns. If the torque is too low, the wheel spins faster than the vehicle moves.

$$\alpha = \frac{a}{t}$$ $$F_{road} = ma$$ $$\tau_{engine} - (F_{road} \times r) = I\alpha$$

• And that would mean that if the wheel were massless, then the torque would be zero and it would not roll, right?
– John
Commented Jan 16, 2020 at 6:58
• Finally an answer that makes sense! Thank you very much! But is there a way to precisely calculate the force at the bottom from engine torque apart from the formula for net torque?
– John
Commented Jan 16, 2020 at 7:01
• Mate any chance you can take a look at this? engineering.stackexchange.com/questions/55295/… You wouldn't believe I've got a Physics degree would you... I was never any good at this mechanics stuff Commented May 19, 2023 at 20:05

Let us consider the particle (or a sum of particles) in contact with the surface. The torque generates force (=60N) on the particle (or sum of particles) in the backward direction or in the direction opposite to the direction in which it intends to move. But the friction acts on the same particle (=60 N) causing the net force on the particle to be 0.

[Please note here that this is not the case of forces on different bodies mentioned in previous answer. This is because the axle producing torque applies a force on the tire, and the friction also applies force on the particle of the tire in contact with it in opposite direction. For example, when pushing a box on a surface with a force less than the max friction, the box will not move as the friction and the force applied by you cancel each other out, both being applied on the box.]

As mentioned above, the particle does have a force = 0N and that is why it acts as an axis of rotation without slipping. The frictional force does not act on the rest of the particles and the torque causes them to rotate about the axis (ie. the particle at rest at that instant) causing it to move forward.

Your logic would have been correct in some cases of linear motion like the case of box mentioned above. In a case like walking, the force by the leg on the floor in the backward direction and the friction acting in the forward direction are applied on different bodies and thus, do not cancel each other out. These two cases should not be confused with one another.

• What happens to the torque? What is it equal to? And how does the friction not act on the whole wheel. What accelerates it then?
– John
Commented Jan 15, 2020 at 16:26
• What do you mean by what happens to the torque? It is equal to 30 N/m. As I said, in linear motion, the friction will act on the whole but in rotation, the part where it acts gets converted to an axis and the torque accelerates the wheel. See it this way- If you ride a bike and pull the front brakes, it would stop the translational (linear) component of the bike. But it would do a wheelie right? That means it is still able to do a rotation even though the brake is applied. The wheel gets converted into an axis. Compare the wheel to the particle and the whole bicycle to the wheel.
– Scar
Commented Jan 15, 2020 at 16:34
• But see that the particle acts as the axis for just an instant. The rotation about that axis causes the point of contact to change to the adjacent particle in the next instance. Thus the car moves as a whole.
– Scar
Commented Jan 15, 2020 at 16:37
• So if I understood you correctly...The torque would be the same 30 Nm, the force that acts on the center of mass would be equal to the friction force which in turn will produce $6 m/s^2$ linear acceleration and because the wheel is not slipping the following is true $\alpha = a / r = 12 rad/s^2$. The moment of inertia of the wheel is $1/2mr^2 = 1.25 kgm^2$ and the torque which is $\tau = \alpha * I = 15 Nm$ and that is half the torque which I started with. What's going on?
– John
Commented Jan 15, 2020 at 16:45
• Why would the force on the center of mass be equal to the frictional force? The friction just converts the particle in contact as a stationary axis for rotation for an instant. The force on the center of mass will be equal to the force produced by the torque. Did you understand the bike analogy?
– Scar
Commented Jan 15, 2020 at 16:49

The only role friction has here is to prevent slipping. It does not influence the linear acceleration. Here is why:

The total KE is equal to the sum of KE of rotation and KE of translational motion. See that both of them are provided by the torque which is generated by the axle. So, the force is provided by the car engine, so to speak.

• And what is "the force provided by the car engine" equal to? Give me concrete numbers for my example. Thanks in advance!
– John
Commented Jan 15, 2020 at 17:09
• 60 N. It has already been covered.
– Scar
Commented Jan 15, 2020 at 17:20
• But that is also the magnitude of the friction force. And the friction force acts in the direction of motion.
– John
Commented Jan 15, 2020 at 17:22
• Your point being?
– Scar
Commented Jan 15, 2020 at 17:28
• You said that the friction force is not the one accelerating the wheel, but the force produced by the wheel itself. But that could not be the case because the force the wheel exerts on the road is in the opposite direction of motion.
– John
Commented Jan 15, 2020 at 17:31

You are making the common mistake of thinking that Newton's third law means that forces should cancel each other resulting in no net force and nothing accelerating.

In order to consider the effect of the equal and opposite forces bodys exert on each other according to Newton's third law, you need to apply Newton's second law to each body individually. Although the magnitude of the forces on the two bodies may be the same, it is the net force on each body individually divided by the mass of each body that determines its acceleration, per Newton's second law.

Now take the example of the wheel. If you draw a free body diagram of the wheel you will see that the only external force acting on the wheel (ignoring other possible dissipative forces) is the static friction force acting forward. That is the net force acting on the wheel and the force responsible for the acceleration of the wheel.

I'm actually more interested about what happens to the torque on the wheel. Is it 0? If not, what is it equal to?

The torque on the wheel is obviously not zero because that would mean the opposing static friction force would also be zero resulting in no acceleration. What happens is the torque on the wheel is transmitted to the ground (the Earth).

A free body diagram of the ground would show the that the clockwise torque of the wheel would be doing work on the earth by causing a counter clockwise torque on the Earth. Technically, that gives the Earth a counter clockwise angular acceleration of

$$α=\frac{τ}{I}$$

Where $$τ$$ is the torque applied to the earth by the wheel, and $$I$$ is the moment of inertia of the Earth. But the moment of inertia of the Earth is so high (about 8 x 10$$^{37} kg.m^2$$) that the angular acceleration is infinitesimal, so it can be ignored.

Hope this helps.

• Comments are not for extended discussion; this conversation has been moved to chat. Commented Jan 15, 2020 at 21:50
user
The passage primarily discusses the physics of a tire accelerating from rest, focusing on the torques and forces involved, and includes a detailed explanation of the role of friction and the distribution of torque between the tire and the ground.

TYPE: NO_QUESTION

The passage does not contain a direct mathematical question that requires a specific numerical answer. Instead, it provides a detailed explanation and discussion of the physics involved in the scenario. While there are calculations and numerical values mentioned, they are used to support the explanation rather than to pose a question.