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The velocity of a vehicle while taking circular turn is tangential to circular path. Friction opposes relative motion between ground and tyres. Then how does friction act radially inwards?

In other words, why does the car tend to go radially outwards although its velocity is tangential?

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Tires/wheels are interesting objects. In the ideal case they allow zero restriction to motion in the plane of the wheel (rolling motion) and they prevent motion perpendicular to the plane of the wheel (sideways motion).

So even though the car is moving forward, the contact patch of the tire and the ground are not moving relative to each other. We presume that friction is zero when the car rolls straight ahead.

When the tire direction is shifted (by turning the steering wheel), a component of the vehicles velocity is now perpendicular to the tire plane. The tire attempts to resist this motion and a friction force appears. Because this force is on the front of the car, it both moves the car sideways and turns the car (and becomes a centripetal force).

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Whenever a body is taking a circular turn, it is being accelerated as it's angle is being changed constantly. At this point, inertia comes into play, or simply a force resisting the acceleration or the change in direction.

This inertial force is acting opposite to the direction of, the change in the direction of motion of the body. While friction, is acting aginst this inertial force, and hence friction is acting radially inwards.

This is the reason during rains, a car easily skids, in a direction opposite to the direction the body is trying to turn. As there is lower friction between the wheels of a car and the ground during rains.

Hope this helps.

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What I get from this question is that the car has a tendency to fall radially outward if the normal reaction of the wheel (that is nearer the center of the turn),becomes zero ie when the inner wheels lift from the ground .If the car had been a smooth body with negligible dimensions, then it would have a tendency to skid tangentially,however overturning of a car is the result of unbalanced torques acting on the caramel not due to the friction acting radially.

You may get this by the example of a formula-1 racing car which has a low center of gravity and a broad base which makes it more stable at high speeds and less susceptible to overturning.Hope it helps.

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Friction force comes in to counterbalance (to certain limit) any force acting on the vehicle. In the radial direction, centrifugal force acts on the vehicle, so there will be friction force in that direction.

Tendency of the car to move radially outwards is because of its inertia (recall Newton's first law of motion).

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We know that whenever a body performs circular motion there has to be a centripetal acceleration which causes this circular motion. The only force that can seemingly provide the centripetal acceleration is frictional force.

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