Consider a point P at the contact point of a | vedclass

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(B) When a wheel of radius $R$ rolls without slipping,the horizontal distance covered by the center of the wheel in half a revolution is equal to $\pi

Vedclass · Accepted Answer

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

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(B) When a wheel of radius $R$ rolls without slipping,the horizontal distance covered by the center of the wheel in half a revolution is equal to $\pi R$.The vertical displacement of the point $P$ (which was initially at the contact point $A$ and moves to the topmost point $A'$) is equal to the diameter of the wheel,which is $2R$.The total displacement is the vector sum of the horizontal and vertical displacements,which is the hypotenuse of a right-angled triangle with sides $\pi R$ and $2R$.Displacement $= \sqrt{(\pi R)^2 + (2R)^2} = R\sqrt{\pi^2 + 4}$.Given $R = 1 \ m$,the displacement $= 1 	imes \sqrt{\pi^2 + 4} = \sqrt{\pi^2 + 4} \ m$.

Consider a point $P$ at the contact point of a wheel on the ground,which rolls on the ground without slipping. Find the displacement of point $P$ when the wheel completes half of a rotation (given the radius of the wheel is $1 \ m$).

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