Question

A wheel of radius 1 meter rolls forward half a revolution on a horizontal ground. The magnitude of the displacement of the point of the wheel initially in contact with the ground is,
A. $2\pi$
B. $\sqrt 2 \pi$
C. $\sqrt {{\pi ^2} + 4}$
D. $\pi$

Answer 1

The above problem can be resolved using the kinematics of rotational motion for any object. Moreover, the concept of displacement covered by any object during is rotational motion is to be keenly analysed. Here, in order to make the complete solution, some specific conditions for the rolling object need to be applied.

Complete step by step answer:
The radius of the wheel is, $R = 1\;{\rm{m}}$.
The horizontal distance covered by the wheel in the half revolution is, $D = \pi R$
Then, the displacement of the point at which the initial contact is made by the wheel with the ground is given as,

{D_1} = \sqrt {{D^2} + 2{R^2}} \\\ {D_1} = \sqrt {{{\left( {\pi R} \right)}^2} + 2{R^2}} \\\ {D_1} = R\sqrt {{\pi ^2} + 4} \end{array}$$ Therefore, the required magnitude of displacement is $$R\sqrt {{\pi ^2} + 4} $$ **So, the correct answer is “Option C”.** **Note:** To resolve the given problem, one must remember the mathematical relation for the horizontal distance covered by any object in its linear path. Moreover, the condition for the displacement caused by the moving object at the time instant, when it makes some initial contact with the ground is to be noted and remembered.