A wheel of radius 2 cm is at rest on a horiz | vedclass

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(A) Let the radius of the wheel be $R = 2 \ cm$. Initially,point $P$ is at the origin $(0, 0)$. After half a rotation,the center of the wheel moves fo

Vedclass · Accepted Answer

$2(\pi^{2}+4)^{1/2} \ cm$

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(A) Let the radius of the wheel be $R = 2 \ cm$. Initially,point $P$ is at the origin $(0, 0)$. After half a rotation,the center of the wheel moves forward by a distance equal to half the circumference,which is $\pi R$. The new position of the center is $(\pi R, R)$. After half a rotation,point $P$ moves from the bottom to the top of the wheel. The new coordinates of point $P$ will be $(\pi R, 2R)$. The displacement $d$ is the distance between the initial position $(0, 0)$ and the final position $(\pi R, 2R)$. Using the distance formula: $d = \sqrt{(\pi R - 0)^2 + (2R - 0)^2} = \sqrt{\pi^2 R^2 + 4R^2} = R\sqrt{\pi^2 + 4}$. Substituting $R = 2 \ cm$: $d = 2\sqrt{\pi^2 + 4} \ cm$ or $2(\pi^2 + 4)^{1/2} \ cm$.

$A$ wheel of radius $2 \ cm$ is at rest on a horizontal surface. $A$ point $P$ on the circumference of the wheel is in contact with the horizontal surface. When the wheel rolls without slipping on the surface,the displacement of point $P$ after half a rotation of the wheel is:

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