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Suppose, a cyclist is moving on a path. Now, static friction can provide the necessary centripetal force when the cyclist wants to take a turn.

However, if the static friction can't provide the necessary centripetal force, the cyclist has to bend and get the necessary centripetal force as the reaction force from the ground (cyclist pushes on the ground and the ground pushes on the cyclist at an angle).

My question is, while the cyclist is taking a turn by bending, does the static friction force still contribute to the centripetal force, or does the horizontal component of the reaction force from the ground only act as the centripetal force?

This might help you in answering the question.

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  • $\begingroup$ "My question is, while the cyclist is taking a turn by bending, does the static friction force still contribute to the centripetal force". Not only does it contribute to the centripetal force, it is the centripetal force. $\endgroup$
    – Bob D
    Nov 17, 2021 at 23:11
  • $\begingroup$ @BobD If the reaction force that the cyclist is getting from the ground is $R$ and the coefficient of static friction is $\mu$, and $\theta$ is the angle that the cyclist is making with the vertical plane while bending towards the center of the circular path, then the centripetal force while bending will be $$R\sin\theta+\mu mg=\frac{mv^2}{r}$$ If the cyclist wasn't bending, then the centripetal force would've been, $$\mu mg=\frac{mv^2}{r}$$ $\endgroup$ Nov 18, 2021 at 3:14
  • $\begingroup$ $m$=mass of the cyclist+cycle, $g$=acceleration due to gravity, $v$= linear velocity of the cyclist while taking the turn. Is this what you mean sir? $\endgroup$ Nov 18, 2021 at 3:16
  • $\begingroup$ $\mu mg$ is not the static friction force. It is the maximum possible static friction force before skidding occurs. The static friction force equals the centripetal force up until the maximum possible static friction force is reached $\endgroup$
    – Bob D
    Nov 18, 2021 at 8:09
  • $\begingroup$ @BobD I understand sir. However, is my explanation correct? $\endgroup$ Nov 18, 2021 at 13:55

2 Answers 2

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The road exerts a force on the tire. Part of this is because the road is a rigid object. The bonds between atoms in the road hold atoms in place. If a tire pushes downward on the road, the atoms push back just hard enough to prevent the tire from penetrating the surface. This is the reaction force.

If a tire pushes sideways and tries to drag road atoms along with it, the atoms push back just hard enough to prevent this, or hard enough that they stay in place while the tire skids. This is static friction. Static because the bottom of the tire does not skid. A patch of tire has $0$ velocity for the moment it has contact.

If the road is horizontal, the sideways force is friction.

If the road is banked, the sideways force is partly the horizontal component of friction and partly the horizontal component of the reaction force.

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Suppose, a cyclist is moving on a path. Now, static friction can provide the necessary centripetal force when the cyclist wants to take a turn.

On an unbanked curve, static friction is the necessary centripetal force as it is the only horizontal force acting towards the center of the turn.

However, if the static friction can't provide the necessary centripetal force, the cyclist has to bend and get the necessary centripetal force...

Again, on a flat surface, the static friction force has to provide the necessary centripetal force as it is the only horizontal force acting on the wheels of the bicycle. The cyclist has to lean into the turn in order not to fall away due to its inertia, as discussed below.

My question is, while the cyclist is taking a turn by bending, does the static friction force still contribute to the centripetal force, or does the horizontal component of the reaction force from the ground only act as the centripetal force?

For a flat (non banked) turn, the horizontal component of the reaction force is the static friction force, i.e., they are one and the same.

The figures below show both the physical and inertial (centrifugal) forces acting on the bike in the rotating (non inertial) reference frame of the turn. Note that the static friction force, $F_{f}=\frac{mv^2}{r}$, is equal and opposite to the centrifugal force in the rotating frame.

The figure at the left shows the forces if the cyclist attempted to make the turn without leaning. Note that there would be a net clockwise torque about the point of contact between the tire and the road due to the inertial force. This would cause the cyclist to fall away from the turn.

The figure to the right shows the forces acting on the cyclist when leaning at an angle $\theta$. The shift of the center of gravity to the left produces a counterclockwise torque about the tire contact point which can now balance the clockwise torque of the inertial force.

From the figure to the right, the leaning angle as a function of the speed and radius of curvature is

$$\tan\theta=\frac{\frac{mv^2}{r}}{mg}$$

$$\theta=\arctan \biggl (\frac{v^2}{rg}\biggr )$$

Note that for a given radius, the greater the cyclist's speed the more the cyclist needs to lean into the turn (the greater $\theta$).

The maximum speed for a given radius is limited by the maximum possible static friction force of $\mu_{s}mg$ where skidding is impending, or

$$\mu_{s}mg=\frac{mv_{max}^2}{r}$$

$$v_{max}^{2}=\mu_{s}rg$$

Substituting for $v_{max}^2$ in the second equation,

$$\theta_{max}=\arctan \mu_s$$

Where $\theta_{max}$ is now the maximum leaning angle without going into a skid.

With respect to the figure to the right, taken from the following Wikipedia article

https://en.wikipedia.org/wiki/Bicycle_and_motorcycle_dynamics

the article points out that it is an idealized case. It discusses its limitations with respect to the dynamics of real bicycles and motorcycles.

Hope this helps.

enter image description here

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