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It is a common observation that riding a bicycle with an inflated tyre is easier than riding one with a deflated tyre but why is it so?

As per my knowledge in an ideal case of no deformation in tyre(when it is inflated) the torque of normal is zero(almost zero in non ideal case) about the axis of rotation whereas in the deflated case the tyre gets deformed and the normal shifts to the front of the tyre and therefore there is a torque of the normal too that our muscles need to overcome.

Is what I thought right and is there any other reason too?

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    $\begingroup$ When the tyre is inflated, the pressure inside the tyre is reducing the normal force between tyre and the ground, and thus reducing the friction between the tyre and the road. Also when the tyre gets deflated, it gets deformed during the rotation of the tyre, the energy you supply is used up in changing the shape and not overcoming the friction, and due to deformation, friction increases. The torque of the normal may shift to the front to cancel the increased amount of friction. But that won't be a major shift. Just keeping all facts to be considered here. I don't know the answer for sure. $\endgroup$ – Quark Jan 13 '16 at 12:29
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    $\begingroup$ @Quark I do not think that the normal will change because if we consider the (bicycle + rider) system then the force due to pressure becomes an internal force and then the mg of the system is balanced lets assume by a normal of N. Now when the tyre deflates, even if the pressure changes, the mg still remains the same (except a minimal change of the mass of air) thus making the N remain almost the same. Now as the normal does not change, if there is slipping(which usually is in practical cases) the kinetic friction does not change much. $\endgroup$ – Pranav Rastogi Jan 13 '16 at 12:37
  • $\begingroup$ @PranavRastogi As you said air has mass too. Suppose a balloon, inflated or deflated, normal would act the same on it. i.e. Perpendicular to point of contact. But which one do you suppose will be easier to roll on the floor? It is rolling friction which is to be considered here and not kinetic friction. I'm not saying I disagree with what you are saying. $\endgroup$ – Quark Jan 13 '16 at 12:53
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There are several factors that may be taken in account, but the more important is the energy used deforming the tire.

Suppose a deflated tire. As you move forward and the tire rotates, the part of the tire that is starting to touch the ground has to be deformed (since the tire is flat). You have to use an important amount of energy for that. Note that the part of the tire that has just stopped touching the ground also has to be deformed, which recovers some energy. Nonetheless, not all energy is gained due to elastic hysteresis, so we have a net loss. That loss has to be even by the cyclist, which is why it is more difficult to ride a bike with flat tires.

Note that there are other factors that may influence this. I'm quoting Wikipedia here

Additional contributing factors include wheel diameter, speed, load on wheel, surface adhesion, sliding, and relative micro-sliding between the surfaces of contact. The losses due to hysteresis also depend strongly on the material properties of the wheel or tire and the surface.

You may want to read this article about rolling resistance. Here it suppose a solid (or perfectly inflated) tire and an elastic ground, but it is equivalent.

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  • $\begingroup$ Agreed, plus cornering is much more difficult because the slack rubber deforms like crazy in response to lateral force. $\endgroup$ – Carl Witthoft Jan 13 '16 at 15:21

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