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Say you have bicycle tire moving with constant velocity. Since the point of contact with the surface remains stationary, so tire will not slide even on an icy surface. But if pushed tangentially to it or in any other direction so as to provide a net torque in its initial direction of travel. It will result in acceleration (translational) of the center of mass as well as an angular acceleration to the whole wheel resulting a tangential (linear) acceleration to the point of contact also. So to prevent sliding of the wheel the force friction must act in the forward direction(i.e opposite to the intended linear one). That sounds quite reasonable.

Diagram of the wheel on the plane surface

But on working with ramps. Say a an initially stationary wheel accelerated due to the force of gravity along the ramp. The force friction acts opposite to the direction of the travel of the CM.

Diagram of a wheel on a plane surface

  • Why there is a change in way the frictional force vector is assigned? ( Does it has something to do with torques and moment of arms.)

  • What if in the former case (horizontal) instead of tangential force the force is due to internal sources (say a bowling ball with gyroscope sort of thing)?


Image Courtesy : Fundamentals of Physics by Resnick, Halliday and Walker

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2 Answers 2

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The force vector of static friction is reversed because, in the first diagram the tire pushes against the road. In the second diagram the road pushes against the tire. Think of it as pedaling the bike on level ground, verses using the brakes going downhill, to maintain steady speed in both cases. Reversing the torque on the wheel, reverses the static friction vector.

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Nothing is changed; the second illustration is just slightly more detailed with how the frictional force is derived.

Why there is a change in way the frictional force vector is assigned? ( Does it has something to do with torques and moment of arms.)

Essentially you have a coefficient of static friction $\mu_S$ between the tire and the surface, which can apply up to $\mu_S F_N$ newtons of resistance to prevent the wheel from slipping, where $F_N$ is the normal force of the surface perpendicularly resisting the weight of your object at the point of contact.

So, in the case of your wheel (which weighs $F_g $), the normal force at the point of contact for a surface ramped at $\theta$ from horizontal will be $$F_N = F_g\cos\theta$$ and so the (maximum) static friction will be $$f_s = \mu_s F_g\cos\theta$$

oriented, of course, in the direction opposite of the motion of the center of the wheel (for the condition to roll without slipping)

What if in the former case (horizontal) instead of tangential force the force is due to internal sources (say a bowling ball with gyroscope sort of thing)?

In your examples, to roll without slipping, your center of mass must travel at $$ \vec{v}_m = \vec{r} \times \vec {\omega}$$ (where $\vec{r}$ is the radial vector from the center to the point of contact, and $\vec{\omega}$ is the angular velocity vector).

A bowling ball sliding along in the forward direction with some back- or side-spin, for example, would not meet that condition.

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