A disc is performing pure rolling on a smooth stationary surface...

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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Options

Avelocity is v, acceleration is zero
Bvelocity is zero, acceleration is zero
Cvelocity is v, acceleration is $\mathrmv^{2} / \mathrm{R}$
Dvelocity is zero, acceleration is $\mathrmv^{2} / \mathrm{R}$

Correct answer

D. velocity is zero, acceleration is $\mathrmv^{2} / \mathrm{R}$

Step-by-step solution

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)

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