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 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
$2\pi $
$\sqrt 2 \pi $
$\sqrt {{\pi ^2} + 4} $
$\pi$
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
Horizontal distance covered by the wheel in half revolution = $\pi R.$
So the displacement of the point which was initially in contact with ground $= AA'$ $ = \sqrt {{{(\pi R)}^2} + {{(2R)}^2}} $
$=R\sqrt {{\pi ^2} + 4} $ $ = \sqrt {{\pi ^2} + 4} $ $(As\,R = 1m)$
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