A point P on the circumference of a wheel is | vedclass

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(D) Let the radius of the wheel be $R = 5 \, m$. After half a revolution,the wheel moves forward by a horizontal distance equal to half its circumfere

Vedclass · Accepted Answer

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

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(D) Let the radius of the wheel be $R = 5 \, m$. After half a revolution,the wheel moves forward by a horizontal distance equal to half its circumference,which is $\pi R$. The vertical displacement of point $P$ is equal to the diameter of the wheel,which is $2R$. The total displacement $d$ is the straight-line distance between the initial and final positions of point $P$,given by the Pythagorean theorem: $d = \sqrt{(\pi R)^2 + (2R)^2} = R\sqrt{\pi^2 + 4}$. Substituting $R = 5 \, m$,we get $d = 5\sqrt{\pi^2 + 4} \, m$.

$A$ point $P$ on the circumference of a wheel is initially in contact with the ground. If the radius of the wheel is $5 \, m$,what will be the displacement of point $P$ when the wheel rolls forward by half a revolution?

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