question_answer

A major-cyclist moving with a velocity of 72 km/h on .1 flat road lakes a turn on the road at a point where the radius of curvature of the road is 20 m. The acceleration due to gravity is 10 m/s2. In order 10 avoid sliding. he must not bend with respect to the vertical plane by an angle greater than :

A) \[\theta ={{\tan }^{-1}}(4)\]                       

B) \[\theta ={{\tan }^{-1}}(25,92)\]

C) \[\theta ={{\tan }^{-1}}(2)\]                       

D) \[\theta ={{\tan }^{-1}}(6)\]

Correct Answer: C

Solution :

The figure shows the rider taking a turn at an angle \[\theta \] with the vertical. Let m be mass of rider and motorcycle, v the speed of motorcycle and r the radius of circular turn.     When the motorcyclist is turned, A  frictional force \[f\left( =\frac{m{{v}^{2}}}{r} \right)\]towards the  centre of the turn is exerted at the point A on the road. Let two equal and opposite forces \[{{F}_{1}}\]and \[{{F}_{2}}\]each equal and parallel to F acts at the centre of gravity G of the rider. Let the rider lean through an angle \[\theta \] from vertical balancing the two couples. Couple formed by mg and R =  couple formed by \[{{F}_{1}}\] and \[{{F}_{2}}\] \[mg\times GA\sin \theta =F\times GA\cos \theta \] \[\tan \theta =\frac{F}{mg}\] But \[F=\frac{m{{v}^{2}}}{r}\] \[\therefore \]     \[\tan \theta =\frac{{{v}^{2}}}{rg}\] Given, \[v=72\,km/h=20\,m/s,\,\,r=20\,m,\] \[g=10\,m/{{s}^{2}}\] \[\therefore \]    \[\tan \theta =\frac{{{(20)}^{2}}}{20\times 10}=2\] \[\therefore \]         \[\theta ={{\tan }^{-1}}(2)\]