Question

A wheel of circumference $C$ is at rest on the ground. When the wheel rolls forward through half a revolution, then the displacement of initial point of contact will be:

• A. $C\sqrt{\frac{1}{{\pi }^2}+\frac{1}{4}}$
• B. $\frac{C}{2}$
• C. $\pi \sqrt{C^2+4}$
• D. $C\sqrt{\frac{1}{\pi }+\frac{1}{2}}$

Answer 1

The correct option is A. $C\sqrt{\frac{1}{{\pi }^2}+\frac{1}{4}}$

Horizontal displacement of the point is half the circumference.

$x=\frac{C}{2}$

Vertical displacement of the point is equal to the diameter.

$y=\frac{C}{\pi }$

Total displacement of the point is:
$d=\sqrt{x^2+y^2}$

$=C\sqrt{\frac{1}{{\pi }^2}+\frac{1}{4}}$