A particle on the outer surface of a wheel,wh | vedclass

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Question

(D) When the wheel completes half a rotation,the particle moves from point $P$ to the topmost point $P'$.The horizontal distance covered by the center

Vedclass · Accepted Answer

$5 \sqrt{\pi^2 + 4} \ m$

Vedclass · Answer

(D) When the wheel completes half a rotation,the particle moves from point $P$ to the topmost point $P'$.The horizontal distance covered by the center of the wheel is $\pi R$,so the horizontal displacement of the particle is $\Delta x = \pi R$.The vertical displacement of the particle is the diameter of the wheel,$\Delta y = 2R$.The total displacement is the vector sum of the horizontal and vertical displacements:$S = \sqrt{(\Delta x)^2 + (\Delta y)^2} = \sqrt{(\pi R)^2 + (2R)^2} = R \sqrt{\pi^2 + 4}$.Given $R = 5 \ m$,the displacement is $S = 5 \sqrt{\pi^2 + 4} \ m$.

$A$ particle on the outer surface of a wheel,which is initially at rest,is in contact with the ground at point $P$. Find the displacement of this particle when the wheel completes half a rotation in the forward direction. (Radius of the wheel $= 5 \ m$)

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