# Torque and Acceleration

1. Feb 1, 2009

I have a bike that has a mass of 100kilograms and its' wheels are 0.722meters in diameter. I am trying to calculate how much torque I would need to accelerate the bike at 1m/s^2.
Can someone please verify my process?

F=ma, F=(100kg)*(1m/s^2)=100Newtons.
T=R*F. T=(0.361m)*(100N)=36.1N*m of torque, right?

2. Feb 2, 2009

### tiny-tim

Looks fine.

3. Feb 2, 2009

allright, so then my question is, at what point on the wheel is this force exerted? Is it exerted on the outside of the wheel where the tire is, or is it exerted on the inside hub of the rim?

4. Feb 2, 2009

### jambaugh

The force on the wheel pushing the whole bike forward is coming from the ground acting via friction on the outside of the wheel. This is why if you spin your tires (say on a grease slick) you won't go anywhere.

5. Feb 2, 2009

allright,just to make sure, so then the torque i calculated in that equation above is the force that is exerted on the OUTSIDE of the wheel right?

6. Feb 2, 2009

### rcgldr

Both. You have a torque in the forward direction at the hub, related to the tension in the chain times the radius of the wheel's sprocket. You have an opposing torque at the circumference of the wheel, related to the force from the ground times the radius of the tire. If the speed of the bicycle is constant, then these torques cancel each other an the wheel does not experience angular acceleration.

7. Feb 2, 2009

okay, so then you are suggesting that the friction between the outside of the wheel and the ground is opposing the motion of the wheel? (you are suggesting that there is a static coefficient of friction to be broken?)

there is torque on both the outside and the inside hub of the wheel, it's just a matter that the torque on the outside of the wheel is much lesser than that on the hub of the wheel when the force is applied through the hub.

8. Feb 3, 2009

### rcgldr

Yes, if you mean opposing relative motion between the contact patch and the ground (tire slipping).

Under normal circumstances, the torques are nearly equal in magnitude and opposing. When the bicycle is accelerating, the torque from the hub results in a backwards force applied to the ground at the contact patch. This coexists with the ground exerting and equal and opposite forwards force to the contact patch, which tranmits the forward force through the axle, and then the rest of the bicycle and rider. This forward force from the ground is opposed by rollising resistance, aerodynamic drag, and interia of bicycle and rider.

Since the wheel also accelerates when the bicycle accelerates there's a small difference in torque, slightly greater at the hub accounting for the rate of angular acceleration of the wheel, which has a small amount of angular inertia compared to the linear inertia of bicycle and rider.

To summarize:

Newton 3rd law pairs: Chain exerts forward force onto top of hub sprocket, top of hub sprocket exerts backwards force on chain. Tire exerts backwards force onto ground, ground exerts forward force onto tire. These are equal and opposing forces.

Torques: Hub torque = chain tension times hub sprocket radius. Tire torque = ground force times tire radius. Normally the torque are very close in magnitude but opposing (if tire is not spinning).

9. Feb 3, 2009

wait, but how can the magnitude of torque be the same for both the hub and the wheel? the hub and the outside of the wheel act as a lever system. since force is being exerted on the hub, and the hub is approximately 13" away from the outside of the wheel, then the force on the outside of the wheel is less than on the hub. (force exerted on the hub is of a very great magnitude, but short distance, and the force on the outside of the wheel is lesser magnitude, but traveling greater distance)

10. Feb 3, 2009

### twowheelsbg

If tyre exerts backward force onto groung, ground exerts forward force onto tyre as friction, equal and opposing, it means that driving force is somewhere else, friction point is not moving - how coud be the driving force there?

And why is such a division introduced - hub torque and tyre torque ?
Tyre is accelerated by the chain via sprocket, the moment is only one ...
and this moment is presented by pair of forces - one at contact point balanced by friction, and the other at the hub, not balanced and accelerating the bike. Contact point is fixed due to the balanced forces there and acts as leverage point for the torque - so driving force at the hub is calculated as moment divided by the wheel radius. Of cource we omit here rolling friction, air drag and other minor frictions

11. Feb 3, 2009

### rcgldr

If the wheel is rotating at a constant speed, then the net torque on the wheel is zero. If the torques aren't equal and opposing, the wheel experiences angular acceleration.

and if the wheel is rotating at constant speed, then the hub torque (hub force times hub radius) is equally opposed by the wheel torque (tire force times tire radius).

The tire is rolling, so the contact patch (friction point) is moving.

12. Feb 3, 2009

allright, thanks, it makes sense now

13. Feb 4, 2009

### twowheelsbg

Guys, i think you are getting confused with the idea of speaking of two kind of torques - hub and tyre surface. The driving moment is one only equal to chain pull times the rear sprocket radius.
This moment is delivered 100% to the rear wheel, and having the contact patch immobialized by the friction forces, this same moment results in force pushing the hub forward - equal to the moment divaded by it's lever: wheel radius ( rolling friction neglected )

https://www.physicsforums.com/library.php?do=view_item&itemid=39
confirming my words that the contact point is not moving, it is changing all the time while the wheel is rolling normaly without sliding. So there is only rolling friction, much smaller in effect of consuming the driving torque energy than the dinamic linear friction of sliding wheel.

Having no acceleration as another case, means that the driving torque resulting from the chain pull is balanced and neutralised by the rolling friction of the tyres, air drag and other minor odds against the movement forward

14. Feb 4, 2009

### jambaugh

Just to clarify some details, if my highly trained intuition hasn't failed me...

If the rider is accelerating the bike then since the angular acceleration of the wheel is not zero, the counter-torque exerted by the ground on the wheel (inhibiting wheel rotation) is less than the torque exerted by the peddle chain on the sprocket of the wheel (promoting rotation).

Now there's a nice little physics problem: Suppose the rider is accelerating the bike at say 0.2 m/s^2 with rim radius 0.361 m and mass 100kg... what's the counter torque due to the ground and what's the driving torque on the sprocket. Given a 0.05m sprocket what's the chain tension? and so on....

Oh! We need the moment of inertia of the wheel. Hmmmm.... shows why you don't want heavy tires on a bicycle.

15. Feb 4, 2009

### twowheelsbg

James, i can't imagine your presentation,
please explain which forces and how create 'counter torque exerted by the ground'

16. Feb 4, 2009

### jambaugh

OK, sorry I wasn't clearer.

Start with a stationary bike. Let's make it move left to right in the x-direction.
Up will be the y-direction and towards us the z-direction.

As the cyclist begins to accelerate he is peddling to make the wheel turn clockwise which means a torque in the -z direction. (Right hand fist with thumb pointing away from you, fingers roll CW.)

Without friction the wheel would accelerate gaining angular momentum (peeling out) and the bike moves nowhere. Given the friction slowing the rate of wheel acceleration it is exerting a "counter torque" i.e. tending to turn it in a counter-clockwise direction (in this case impeding clockwise acceleration) so that's a torque in the +z direction. (Right hand fist thumb extended toward you fingers curl in CCW direction.)

This torque is due to the force acting by the ground on the tread and this force is also pushing the bike forward.

Finally note that though the torques are almost equal in magnitude and opposite. The tire does have some angular acceleration (not as much if it slid freely), it gains angular momentum as the bike accelerates forward and so the two torques can't quite cancel out... the sprocket torque must be slightly larger in magnitude because there is a net angular acceleration in the clockwise direction.

Figure the angular acceleration times the moment of inertia of the tire and this is the net torque.

Once the bike is up to speed then no more acceleration so the torques exactly cancel. As mentioned they correspond to a force from the ground just enough to overcome wind resistance. (This assumes a frictionless bearing in the wheel... if not then there's a third frictional ("counter") torque acting in the +z direction which must be considered).