A disc is performing pure rolling on a smooth | vedclass

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(D) The correct option is $(d)$.As the disc is in combined rotation and translation,each point on the disc has a translational velocity $v$ (in the fo

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velocity is zero,acceleration is $v^2/R$

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(D) The correct option is $(d)$.As the disc is in combined rotation and translation,each point on the disc has a translational velocity $v$ (in the forward direction) and a tangential velocity $v = R\omega$ (due to rotation).For the lowermost point of the disc:$1$. The net velocity is $v_{	ext{net}} = v - R\omega$. Since it is pure rolling,$v = R\omega$,therefore $v_{	ext{net}} = v - v = 0$.$2$. The acceleration of any point on a rotating body has two components: tangential acceleration $(a_t)$ and centripetal acceleration $(a_c)$.- Since the angular velocity $\omega$ is constant,the tangential acceleration $a_t = R\alpha = 0$.- The centripetal acceleration is $a_c = v^2/R$ directed towards the center of the disc.- Thus,the net acceleration of the lowermost point is $a = a_c + a_t = v^2/R + 0 = v^2/R$,directed towards the center of the disc.

$A$ disc is performing pure rolling on a smooth stationary surface with constant angular velocity as shown in the figure. At any instant,for the lowermost point of the disc -

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