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In the section 3.7.3 (Banking of road) we have not included the friction exerted by the road on the car. Suppose the coefficient of static friction between the car tyre and the surface of the road is mu_s, calculate the minimum speed with which the car can take safe turn? When the car takes turn in the banked road, the following three forces act on the car. (1) The gravitational force mg acting downwards (2) The normal force N acting perpendicular to the surface of the road (3) The static frictional force f acting on the car along the surface. |
Answer» Solution : The following figure shows the forces acting onthe horizontal and vertical direction. When the car takes turn with the SPEED v, the centripetal force is exerted by horizontal component of normal force and static frictionalforce. It is given by `N sin theta + f cos theta = (mv^2)/(r)`..(1) In the vertical direction, there is no acceleration. It implies that the vertical component of normal force is balanced by downward gravitational force and downward vertical component of frictional force. This can be expressed as `N cos theta - mg + sin theta" or" N cos theta - f sin theta=mg `...(2) Dividing the equation (1) by equation (2), we get `(N sin theta + f cos theta)/(N cos -f sin theta) = (v^2)/(rg)`...(3) To calculate the maximum speed for the safe turn, we can use the maximum static friction is given by` f= mu_s N` . By substituting this relation in equation (3), we get `(N sin theta+ mu_s N cos theta)/(N cos theta - mu_s N sin theta ) = (v_(max)^2)/(rg)` By simplifying this equation `(N cos theta {(N sin theta)/(N cos theta) + mu_s})/(N cos theta ( 1- mu (N sin theta)/(N cos theta) ))= (v_(max)^2)/(rg)` ` ((tan theta + mu_s))/(1 - mu_s tan theta) = (v_(max)^2)/(rg)` The Maximum speed for safe turn is `v_(max) = SQRT( rg ((tan theta + mu_s)/(1 -mu_s tan theta)) `...(4) Suppose we neglect the effect of friction (`mu_s`= 0), then safe speed `v_(safe)= sqrt( rg tan theta) `...(5) Note that the maximum speed with which the car takes safe turn is increased by friction (equation (4)). Suppose the car turns with speed `v lt v_(safe)` , then the static friction acts up in the slope to prevent from inward SKIDDING. If the car turns with the speed little greater than, then the static friction acts down the slope to prevent outward skidding. But if the car turns with the speed greater than `v_(safe)`then static friction cannot prevent from outward skidding. |
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