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Dhivya was a talented cyclist. She used to go to school in her bicycle presented by her Dad. She used to ride faster, as she always goes late from home. When she was going by the road one day. She skid down from her cycle and got injured. That was a curved road where she has to take a turn, and as she was in hurry, this incident has happened. Why does Dhivya skid from her cycle ? (i) What is the condition for skidding ? (ii) Derive an equation for the Angle for safer riding. |
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Answer» Solution :(i) While moving in a circular level road of radius 'r' at a velocity 'v' a cyclist has to bend by an angle `theta` from the vertical given by the expression `tan theta = (v^(2))/(rg)` to stay in equilibrium (i.e. To AVOID a skidding or fall). So Dhivya while taking a circular path, must have gone with a particular speed to PREVENT skidding. Condition for skidding When the CENTRIPETAL force is greater than the frictional force. Skidding occurs If `mu` is the co-efficient of fircition between the road and tyre then the limiting friction (firctional force) is f = `mu`R where Normal reaction R = mg , f = `mu`(mg) Thus for skidding, Centripetal force `gt` frictional force `(mv^(2))/(r) gt mu (mg) , (v^(2))/(rg) gt mu` `(v^(2))/(rg) = tan theta (theta-"Angle of Banking")` `therefore tan theta gt mu` (In this condition, skidding occurs) (ii) Let r be the radius of circular path m be the mass of Dhivya (cyclist) along with the bicycle, v be the velocity When the cyclist takes the curve she bends inwards from the vertical making an angle `'theta'`. Let 'R' be the reaction of the ground an Dhivya. R can be resolved into two components (i) `R sin theta rarr` acting towards the centre of curve Providing necessary centripetal force. (ii) `R cos theta rarr` balancing the weight of the cyclist along with the bicycle. `R sin theta = (mv^(2))/(r)""...(1)` `R cos theta = mg""...(2)` DIVIDING (1) by (2), `(R sin theta)/(R cos theta) = ((mv^(2))/(v))/(mg)` `tan theta = (v^(2))/(rg) , theta = tan^(-1) ((v^(2))/(rg))` |
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