# Traction force on tire

my sprocket is rigidly attached to rear axel ,is there any difference in torque at sprocket and tire?

Chestermiller
Mentor
What do you get if you apply an angular momentum balance to the wheel?

What do you get if you apply an angular momentum balance to the wheel?

Or more basically, suppose you have some torque at the sprocket. Would you have to exert more or less rotating force at the tire to create that same torque at the sprocket? Remember that torque = force times radius.

Or more basically, suppose you have some torque at the sprocket. Would you have to exert more or less rotating force at the tire to create that same torque at the sprocket? Remember that torque = force times radius.
less bcz torque =force*radius and in this case sprocket radius is less than that of wheels

Chestermiller
Mentor
The sprocket torque minus the tire torque is equal to the wheel moment of inertia times the angular acceleration of the wheel. If the bike is not accelerating, then the sprocket torque is equal to the wheel torque. (I assume that by sprocket torque, you are referring to the torque applied by the chain to the sprocket).

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Exactly. The farther you get from the center of rotation the less the torque. So the torque at the sprocket is not the same as the torque at the tire.

Chestermiller
Mentor
Exactly. The farther you get from the center of rotation the less the torque. So the torque at the sprocket is not the same as the torque at the tire.
If the car isn't accelerating, it sure is the same. The moments (torques) must balance. Just take the wheel and attached sprocket as a free body.

jack action
Gold Member
$$T_{tire} = T_{sprocket} - I\alpha$$
Where $I$ is the inertia of the sprocket-axle-wheel-tire assembly and $\alpha$ is the rotational acceleration of the assembly.

If there is no acceleration ($\alpha = 0$), then the tire torque is exactly the same as the sprocket torque. Otherwise, in typical assembly, the inertia should be small enough to consider that they are both practically the same as well (although, not necessarily a negligible difference).

Chestermiller
cjl
Exactly. The farther you get from the center of rotation the less the torque. So the torque at the sprocket is not the same as the torque at the tire.
The torque is absolutely the same. The force, however, is not, since as you pointed out earlier, torque is force multiplied by radius.

Chestermiller
The sprocket torque minus the tire torque is equal to the wheel not accelerating, then the sprocket torque is equal to the wheel torque. (I assume that by sprocket torque, you are referring to the torque applied by the chain to the sprocket).
power=torque*angular velocity , since p
less bcz torque =force*
The torque is absolutely the same. The force, however, is not, since as you pointed out earlier, torque is force multiplied by radius.
hmm ,power remains constant i.e power =torque *angular velocity ,angular velocity for sprocket and wheel is same so torque nust be equal. Thank u all.

billy_joule
hmm ,power remains constant i.e power =torque *angular velocity ,angular velocity for sprocket and wheel is same so torque nust be equal. Thank u all.

That is not always true, there can be losses between the sprocket and the wheel; friction at a bearing, losses through a CV joint, a dragging brake pad etc etc. Power still remains constant but not all of it makes it to the wheel:
Psprocket ≠ PWheel

Psprocket = PWheel + Pbearing losses PCV losses + Pany other losses

Whether these losses are small enough to be ignored depends on your system.

Aashish sarode