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(A) When the disc rolls without slipping,the center of the disc moves forward by a distance equal to the arc length covered. For half a rotation,the c

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) When the disc rolls without slipping,the center of the disc moves forward by a distance equal to the arc length covered. For half a rotation,the center moves by a horizontal distance of $\pi R$.The point $A$,which was initially at the top,moves to the bottom of the disc after half a rotation. The vertical displacement of point $A$ is equal to the diameter of the disc,which is $2R$.The horizontal displacement of point $A$ is equal to the distance moved by the center,which is $\pi R$.The total displacement $d$ is the vector sum of the horizontal and vertical displacements:$d = \sqrt{(	ext{horizontal displacement})^2 + (	ext{vertical displacement})^2}$$d = \sqrt{(\pi R)^2 + (2R)^2}$$d = \sqrt{\pi^2 R^2 + 4R^2}$$d = R \sqrt{\pi^2 + 4}$

$A$ disc is rolling without slipping on a surface. The radius of the disc is $R$. At $t=0$,the top most point on the disc is $A$ as shown in the figure. When the disc completes half of its rotation,the displacement of point $A$ from its initial position is

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