A cyclist on the ground goes round a circular path of circumference 34.3m in \[\sqrt {22} \] sec. The angle made by him with the vertical will be,
A. 45°
B. 40°
C. 42°
D. 48°

Hint:When a cyclist is moving around a circular path than the center of mass shifts towards the center of the circular path. To balance the moment of force the angle with the vertical be such that the component of the weight of the cyclist is balanced by the outward force along the inclined line.

Formula used:
\[{f_0} = \dfrac{{m{v^2}}}{r}\]
Here \[{f_0}\] is the radially outward force of mass m moving in a circular path of radius r with speed v.

Complete step by step solution:

Image: Free body diagram of cyclist


Let the angle made by the cyclist vertically is \[\theta \]. At equilibrium, the net force acting on the cyclist must be zero. We resolve the force along the inclined line. Let the inclined plane be the x-axis with an upward direction as positive and a downward direction as negative. The net force acting on the cyclist along the x-axis must be zero to attain equilibrium while making the circular path.
\[\sum {{F_x}} = 0\]
\[\Rightarrow \dfrac{{m{v^2}}}{r}\sin \theta - mg\cos \theta = 0\]

By simplifying, we get
\[\dfrac{{m{v^2}}}{r}\sin \theta - mg\cos \theta = 0\]
\[\Rightarrow \dfrac{{m{v^2}}}{r}\sin \theta = mg\cos \theta \]
\[\Rightarrow \tan \theta = \dfrac{{rg}}{{{v^2}}}\]
\[\Rightarrow {\tan ^{ - 1}}\left( {\dfrac{{rg}}{{{v^2}}}} \right)\]
If the radius of the path is r, then using the circumference formula,
\[2\pi r = C\]
\[\Rightarrow r = \dfrac{C}{{2\pi }}\]

The speed is the distance travelled per unit of time,
\[v = \dfrac{C}{t}\]
Putting the values, we get
\[\theta = {\tan ^{ - 1}}\left( {\dfrac{{\left( {\dfrac{C}{{2\pi }}} \right)g}}{{{{\left( {\dfrac{C}{t}} \right)}^2}}}} \right)\]
\[\Rightarrow \theta = {\tan ^{ - 1}}\left( {\dfrac{{g{t^2}}}{{2\pi C}}} \right)\]
\[\Rightarrow \theta = {\tan ^{ - 1}}\left( {\dfrac{{9.8 \times {{\left( {\sqrt {22} } \right)}^2}}}{{2\pi \times 34.3}}} \right)\]
\[\Rightarrow \theta = {\tan ^{ - 1}}\left( {1.00} \right)\]
\[\therefore \theta = 45^\circ \]
Therefore, the required angle made by the cyclist with the vertical is 45°.

Therefore, the correct option is A.

Note: When the road is frictional then the frictional force acts opposite to the direction of potential skidding of the tyre of the cycle. If the outward force on the cyclist is greater or less than the frictional force between the tyre and the road then the cyclist can lose balance. So to balance the forces cyclists make an angle vertically.