Question

A disc of mass m and radius R is given translational velocity vo on rough horizontal surface. The
velocity of the disc when it rolls without slipping is

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Answer 1

Answer:

Velocity of the disc is when it rolls without slipping is $\dfrac{2v_0}{3}$

Explanation:

Given:

Mass of the disc=m

Radius of the disc=R

initial translational velocity=$v_0$

Using conservation of Energy we have

let v be the final velocity of the disc when it rolling without slipping and$\omega$ be the angualr velocity of the disc

Friction force actin on the disc=$\mu mg$

acceleration of the disc=$\mu g$

The friction is providing torque

Let f be the frictional force then I be the moment of inertia about centre of the mass and $\alpha$ be the angular acceleration of the disc

$fR=I\alpha\\\mu mg=\dfrac{MR^2}{2}\alpha\\\alpha=\dfrac{2\mu g}{R}$

The final velocity of the centre of the mass of the disc

$v=v_0-at\\v=v_0-\mu gt$

The final angular velocity of the disc

$\omega=\alpha t\\=\dfrac{2\mu g}{R}t$

When the disc rolls the velocity of the centre point of the contact of disc with the ground is zero it means we have

$v=r\omega$

$v_0-\mu gt=\dfrac{2\mu g}{R}t\times R\\\\t=\dfrac{v_0}{3\mu g}$

The velocity when it rolls without slipping is v given by

$v=v_0-\mu gt\\\\v=v_0-\dfrac{v_0}{3\mu g}\times \mu g\\v=\dfrac{2v_0}{3}$

hence the final velocity is calculated.