0
$\begingroup$

Rolling friction supports forward motion of the rolling body. So acceleration of the body equals friction/mass. While moving with uniform velocity, a=0, so friction should also be 0.

$\endgroup$
1
  • $\begingroup$ Friction is not the only force there, but is acting together with the same ( but direction ) force giving the zero net force. $\endgroup$
    – Poutnik
    Commented May 12, 2022 at 12:49

3 Answers 3

1
$\begingroup$

Friction is the force resisting the relative motion of surfaces sliding (or trying to slide or rolling) against each other. Friction doesn't provide the force to a vehicle to move forward but it makes the moving vehicle to come at rest. The vehicle applies an amount of force on surface diagonally by its wheels and the surface also applies a force diagonally on the vehicle as reaction force. If the horizontal component of the reaction force becomes greater than friction, the vehicle starts moving. When a vehicle is moving with constant velocity on a surface, it has to apply a constant amount of force on ground to overcome its friction and continue moving. Friction never becomes 0 between sliding (or trying to slide or rolling) surfaces against each other. So, a vehicle also experience friction while moving with constant velocity.

$\endgroup$
2
  • $\begingroup$ Actually, when you step on the gas pedal, friction between your tires and the road does provide the force that moves the vehicle forward. If your car were resting on a level, frictionless surface, you would not be able to go anywhere. Anybody who's ever tried to drive on a slick, icy parking lot or road knows what happens if you try to accelerate too hard. $\endgroup$ Commented May 12, 2022 at 13:58
  • $\begingroup$ Also, "Friction never becomes 0 between sliding surfaces...So, a vehicle also experience friction while moving with constant velocity." Under normal driving conditions, your tires don't slide against the road at all. There should only ever be static friction between your tires and the road. When the tires break loose and slide, we call that a skid, and we usually try to avoid it. See also: en.wikipedia.org/wiki/Rolling_resistance $\endgroup$ Commented May 12, 2022 at 14:01
0
$\begingroup$

If the tires are rubber & inflated with air, then every rotation of the wheel will flex and then unflex the rubber, dissipating energy by internal friction.

In addition, when a rubber tire rolls into contact with the pavement, it takes work to push the tire tread down into a firm grip on it- and it then takes work to peel the tread out of contact with the pavement. In both cases, friction will dissipate energy as the gripping/ungripping process takes place.

So a rubber tire will furnish rolling friction which acts to slow down the vehicle and heat up the tire even when the vehicle is not accelerating.

Note that if the tire is insufficiently inflated, the rolling flexure will be of greater amplitude and the heat generated by internal friction in the rubber will be greater as well- which can cause the rubber to heat up so much as to lose its strength and the tire to self-destruct violently in smoke and sometimes flames.

$\endgroup$
0
$\begingroup$

Rolling friction supports forward motion of the rolling body

It is static friction that is responsible for initiating forward motion (accelerating) a rolling object without slipping.

While moving with uniform velocity, a=0, so friction should also be 0.

Assuming there are no external forces opposing motion, such as air resistance, then static friction is not required to keep the object rolling at uniform velocity.

There is, however, the possibility of rolling resistance, which is sometimes referred to as rolling friction, to cause objects such as rubber tires to slow down. Rolling resistance is due to the inelastic deformation the rubber of the tire experiences when it is in contact with the road. It in involves friction heating. See this article on rolling resistance from Wikipedia: https://en.wikipedia.org/wiki/Rolling_resistance

Hope this helps.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.