Question

A uniform disc of mass and radius $R$ is projected horizontally with velocity $v_c$ on a rough horizontal floor so that it starts off with a pure sliding motion at $t=0$. After $t_0$ second it acquires pure rolling motion as shown in figure. Calculate the velocity of the centre of mass of the disc at $t_0{.}^{\prime }{\mu }^{\prime }$ is the coefficeient of friction.

• A. $\frac{2V_0}{3}$
• B. $V_0$
• C. $\frac{3}{2}{V}_0$
• D. $\frac{V_0}{2}$

Answer 1

Answer: (A)

$\frac{2V_0}{3}$

$Given$ :- A uniform disc of mass $m$ and radius $R$

Velocity of projection is $V_0$

$ToFind$ :- velocity of center of mass $\left(V_{cm}\right)$ of disc at $t_0$

$Solution$ :- From the $Ist$ case diagram , we know,

Torque about $p$ $\left({\tau }_p\right)$ = $0$

$⟹$ Angular momentum about $p$ will remain conserved ,

$\therefore$ $l_i=l_F$ $------\left(i\right)$

[Here, $l_i=m{V}_{0}R$ $--------$ from $Ist$ case

And $l_F=\left(\frac{3}{2}m{R}^2\right)\omega$ $-------$ from $IInd$ case]

( By parallel axis theorem)

$\therefore V_0=\left(\frac{3}{2}R\right)\omega$ (applying in $\left(i\right))$

$\omega =\frac{2}{3R}{V}_0$

We know , during rolling , $V$= $\omega R$

$\therefore \underset{–}{V_{cm}=\frac{2}{3}{V}_0}$

Hence , $\underset{–}{OptionA\left(\frac{2}{3}{V}_0\right)iscorrect.}$