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A disc is performing pure rolling on a smooth stationary surface with constant angular velocity as shown in figure. At any instant, for the lower most point of the disc $-$
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The correct answer is:
velocity is zero, acceleration is $\mathrm{v}^{2} / \mathrm{R}$
As the disc is in combined rotation and translation, each point has a tangential velocity and a linear velocity in the forward direction. From figure $\mathrm{v}_{\text {net }}$ (for lowest point) $=\mathrm{v}-\mathrm{R} \omega=\mathrm{v}-\mathrm{v}=0$
and acceleration
$=\frac{\mathrm{v}^{2}}{\mathrm{R}}+0=\frac{\mathrm{v}^{2}}{\mathrm{R}}$
(since linear speed is constant)
and acceleration
$=\frac{\mathrm{v}^{2}}{\mathrm{R}}+0=\frac{\mathrm{v}^{2}}{\mathrm{R}}$
(since linear speed is constant)
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