How is static torque calculated for constant speed?

  • #1

Main Question or Discussion Point

The way I have understood it is that dynamic torque involves angular acceleration, while static torque has zero angular acceleration:

α =Δω/Δt = 0

But static torque still has centripetal acceleration with constant speed:

ac = v2/r

where v is linear velocity, and r is the radius

Then F = mac = (mv2)/r

Torque = F . r
Therefore static torque τs = ( (mv2)/r ) . r = mv2


Let us consider a wheel of mass 10kg powered through a driveshaft connected to the output of an engine/motor, with a linear velocity of 10m/s (36km/h) on a road.

Using the equation above, the static torque at the wheel due to constant speed:
τs = 10.(102) = 1000Nm

1) What is the gap in my understanding? I do not see how a wheel travelling at 10m/s could possibly generate 1000Nm. If we consider freeway speeds: 36m/s (~130km/h). Then the static torque is 12,960Nm.

2) Say we also consider the actual car attached to the wheel which is moving. Do we need to add the mass of the car (1000kg) to the wheel's mass? So at 10m/s:
τs = 1010.(102) = 101,000Nm

I am guessing no! So what gives?
 

Answers and Replies

  • #2
BvU
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But static torque still has centripetal acceleration with constant speed:

ac = v2/r

where v is linear velocity, and r is the radius

Then F = mac = (mv2)/r
Nope. Centripetal acceleration is perpendicular to angular acceleration.
With ##\ \vec \tau = I \vec \alpha\ ## you can see that angular acceleration and torque are aligned, not perpendicular.

Your Fc holds the wheel together. That usually does not involve work.

Vehicles need torque to a) accelerate and b) overcome friction.
 
  • #3
Nope. Centripetal acceleration is perpendicular to angular acceleration.
With ##\ \vec \tau = I \vec \alpha\ ## you can see that angular acceleration and torque are aligned, not perpendicular.

Your Fc holds the wheel together. That usually does not involve work.

Vehicles need torque to a) accelerate and b) overcome friction.
Thank you for your response.
Yes, you are right...I should have considered the direction of the vectors. Centripetal force is pulling inwards and that is not aligned with torque.

I do not understand how there can be no torque at the wheel when it is moving at a constant speed though.

Sure, we need a net torque to accelerate. However there is a force being exerted at the wheels when it is at constant speed right? How do I calculate the torque when there is no angular acceleration?
 
  • #4
BvU
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How do I calculate the torque when there is no angular acceleration?
A moving vehicle experiences a force opposite to the direction of motion. There are various sources: rolling friction, and drag from the wind ( more here ). The latter is dominant at higher speeds and is proportional to the square of the speed.

Also: https://en.wikipedia.org/wiki/Automotive_aerodynamics
https://en.wikipedia.org/wiki/Automobile_drag_coefficient
https://en.wikipedia.org/wiki/Drag_coefficient
 
  • #5
A moving vehicle experiences a force opposite to the direction of motion. There are various sources: rolling friction, and drag from the wind ( more here ). The latter is dominant at higher speeds and is proportional to the square of the speed.

Also: https://en.wikipedia.org/wiki/Automotive_aerodynamics
https://en.wikipedia.org/wiki/Automobile_drag_coefficient
https://en.wikipedia.org/wiki/Drag_coefficient
Interesting, so basically what we are interested in here is fluid friction and rolling friction.

If a car is travelling at constant speed, and we somehow managed to suddenly remove fluid friction and rolling friction - there would be a net torque at the wheels and hence it would begin to accelerate right?

To find the static torque at a constant speed, I would need to know the sum of the opposing forces at the wheel (rolling friction + fluid friction).
That is not trivial, as it would vary considerably depending on tyre material, road surface, air density, and of course shape of the car.

If we know the force acting on the wheel at constant speed, then we can also know the static torque at that point in time?
 
  • #6
CWatters
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I think it's a mistake to think about static and dynamic torque. Just think about the net torque. If the net torque is zero there in no angular acceleration. If the net torque is non zero then there must be angular acceleration.

The engine of a car tries to turn the wheels in one direction. Rolling resistance, drag bearing friction etc act in the opposite direction. If these sum to zero there is no net torque so no acceleration (but the car can still be moving at constant velocity). If one is greater than the other the net torque is non zero and there must be acceleration .

It might help to know that the equations for linear motion and angular motion are similar...

Net force = mass * acceleration
Net Torque = moment of inertia * angular acceleration.

All the other equations of linear motion have angular equivalents.
 
  • #7
CWatters
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If a car is travelling at constant speed, and we somehow managed to suddenly remove fluid friction and rolling friction - there would be a net torque at the wheels and hence it would begin to accelerate right?
Correct. At constant velocity the net torque is zero. If you remove rolling resistance the net torque on the wheels becomes non zero and it accelerates.

To find the static torque at a constant speed, I would need to know the sum of the opposing forces at the wheel (rolling friction + fluid friction).
That is not trivial, as it would vary considerably depending on tyre material, road surface, air density, and of course shape of the car.
Indeed, these are all factors that affect the power required and hence the fuel consumption. Some can be measured in a wind tunnel.

In some cases you can tow the object and measure the tension in the tow rope but that has its own issues.

If we know the force acting on the wheel at constant speed, then we can also know the static torque at that point in time?
Yes you would know the engine torque because they are equal and opposite so they sum to zero.
 

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