A disc is performing pure rolling on a smooth | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

Vedclass · Accepted Answer

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

Vedclass · Answer

(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 -

Similar Questions

In the case of pure rolling,what will be the velocity of point $A$ of the ring of radius $R$?

$A$ sphere of mass $M$ and radius $r$ slips on a rough horizontal plane. At some instant,it has translational velocity $V_0$ and rotational velocity about the centre $\frac{V_0}{2r}$. The translational velocity when the sphere starts pure rolling motion is

$A$ wheel is rolling along the ground with a speed of $2\ m/s$. The magnitude of the velocity of the points at the extremities of the horizontal diameter of the wheel is equal to:

If a solid sphere is rolling without slipping on a horizontal plane,then the ratio of its rotational and total kinetic energies is

$A$ solid sphere is rolling without slipping on a horizontal plane. The ratio of the linear kinetic energy of the centre of mass of the sphere and rotational kinetic energy is $:$

Vedclass Test Series

Exam Paper Generator

Online Exam Module