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Laws Of Motion

Question
CBSEENPH11017004

A cyclist tends to negotiate a curved track. Obtain an expression for the angle which he will have to make with vertical.

Solution
Consider a cyclist of negotiating a curve of radius r with velocity v.
Weight of the cyclist, W = mg
              
In order to provide the necessary centripetal force, the cyclist leans through angle straight theta to the power of straight o in inward direction as shown in figure above. 

The various forces acting on the cyclist are: 
(i) Weight Mg acting vertically downward at the centre of gravity of cycle and the cyclist. 
(ii) The reaction R of the ground onto the cyclist, acting along a line, making angle straight theta to the power of straight o with the vertical. 
The cyclist while taking the turn is in equilibrium, therefore the vertical component Rcosθ of the normal reaction R will balance the weight of the cyclist.
The horizontal component R sinθ will provide the necessary centripetal force to the cyclist. 
i.e.,        space space space R space cos theta space equals space M g                  ...(1) 
and            R space sin theta space equals space fraction numerator M v squared over denominator r end fraction               ...(2) 
Dividing (2) by (1), we have
                    tan space theta space equals space fraction numerator v squared over denominator r g end fraction 
rightwards double arrow                     theta space equals space tan to the power of negative 1 end exponent open parentheses fraction numerator v squared over denominator r g end fraction close parentheses 
Therefore, the cyclist should lean inwards at angle straight theta given by tan to the power of negative 1 end exponent space open parentheses straight v squared over rg close parentheses