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, \[v=72\,km/h=20\,m/s,\,\,r=20\,m,\] \[g=10\,m/{{s}^{2}}\] \[\therefore \] \[\tan \theta =\frac{{{(20)}^{2}}}{20\times 10}=2\] \[\therefore \] \[\theta ={{\tan }^{-1}}(2)\]You need to login to perform this action.
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