A wheel of radius 1 m rolls through 180^{cir | vedclass

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Question

(C) The distance travelled by the wheel in a half revolution is given by $d = \frac{C}{2} = \frac{2 \pi r}{2} = \pi r$,where $C$ is the circumference

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) The distance travelled by the wheel in a half revolution is given by $d = \frac{C}{2} = \frac{2 \pi r}{2} = \pi r$,where $C$ is the circumference of the wheel.After a half revolution,the point initially in contact with the surface (point $A$) moves to the top of the wheel (point $B$).The horizontal distance covered by the center of the wheel is $\pi r$,and the vertical displacement of the point is equal to the diameter of the wheel,which is $2r$.Using the Pythagorean theorem for the displacement vector $AB$:$AB = \sqrt{(	ext{horizontal distance})^2 + (	ext{vertical distance})^2}$$AB = \sqrt{(\pi r)^2 + (2r)^2} = r \sqrt{\pi^2 + 4}$Given $r = 1 \ m$,the magnitude of the displacement is $AB = \sqrt{\pi^2 + 4} \ m$.

$A$ wheel of radius $1 \ m$ rolls through $180^{\circ}$ over a plane surface. The magnitude of the displacement of the point of the wheel initially in contact with the surface is:

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