Traction force on tire

[Aashish sarode]

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

 

[Chestermiller]

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

 

[Aashish sarode]

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

will you please elaborate your question?

 

[OldYat47]

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.

 

[Aashish sarode]

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]

will you please elaborate your question?

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).

 

[OldYat47]

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]

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]

$$T_{tire} = T_{sprocket} - I\alpha$$
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]

All. Please see the following thread from July 2015: https://www.physicsforums.com/threads/gear-ratio-in-bicycles-using-rotational-motion.824438/

 

[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.

 

[Aashish sarode]

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:
P_sprocket ≠ P_Wheel

P_sprocket = P_Wheel + P_{bearing losses} P_{CV losses} + P_{any other losses}

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