Question

Find the maximum speed at which a car weighing 3000 kg can negotiate a curve of radius 100 m
if the coefficient of friction between the road surface and the tyres is 0.13. To what angle
should the road be banked if the vehicle is to go around the curve at this speed without
depending upon friction? If the road is 5 m wide at the curve, what would be the elevation of
the outer edge above the horizontal plane passing through the inner edge? g= 9.8 m/s?​

Answer 1

Answer:

Any force or combination of forces can cause a centripetal or radial acceleration. Just a few examples are the tension in the rope on a tether ball, the force of Earth’s gravity on the Moon, friction between roller skates and a rink floor, a banked roadway’s force on a car, and forces on the tube of a spinning centrifuge.

Any net force causing uniform circular motion is called a centripetal force. The direction of a centripetal force is toward the center of curvature, the same as the direction of centripetal acceleration. According to Newton’s second law of motion, net force is mass times acceleration: net F = ma. For uniform circular motion, the acceleration is the centripetal acceleration—a = ac. Thus, the magnitude of centripetal force Fc is Fc = mac.

By using the expressions for centripetal acceleration ac from ac=v2r;ac=rω2

, we get two expressions for the centripetal force Fc in terms of mass, velocity, angular velocity, and radius of curvature:  Fc=mv2r;Fc=mrω2

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You may use whichever expression for centripetal force is more convenient. Centripetal force Fc is always perpendicular to the path and pointing to the center of curvature, because ac is perpendicular to the velocity and pointing to the center of curvature.

Note that if you solve the first expression for r, you get r=mv2Fc

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This implies that for a given mass and velocity, a large centripetal force causes a small radius of curvature—that is, a tight curve.

Explanation: