Find the displacement of the point of contact | vedclass

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(B) When a ring of radius $R$ completes a half-rotation,the center of the ring moves forward by a distance equal to half the circumference,which is $\

Vedclass · Accepted Answer

$R\sqrt {{\pi ^2} + 4} $

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(B) When a ring of radius $R$ completes a half-rotation,the center of the ring moves forward by a distance equal to half the circumference,which is $\pi R$.The point of contact,which was initially at the bottom,moves to the top of the ring. The vertical displacement of this point is equal to the diameter of the ring,which is $2R$.The horizontal displacement of the point is $\pi R$.The total displacement is the vector sum of the horizontal and vertical displacements:$	ext{Displacement} = \sqrt{(	ext{Horizontal Displacement})^2 + (	ext{Vertical Displacement})^2}$$	ext{Displacement} = \sqrt{(\pi R)^2 + (2R)^2}$$	ext{Displacement} = \sqrt{\pi^2 R^2 + 4R^2}$$	ext{Displacement} = R\sqrt{\pi^2 + 4}$

Find the displacement of the point of contact with the ground when a ring of radius $R$ completes a half-rotation.

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