Question

A disc of mass $M$ and radius $R$ rolls with out slipping on a horizontal surface. If the velocity of its centre is $v_0$, then the total angular momentum of the disc about a fixed point $P$ at a height $3/2R$ above the centre $C$

• A. increase continuously as the disc moves away
• B. decreases continuously as the disc moves away
• C. is equal to $2MR{v}_0$
• D. is equal to $MR{v}_0$

Answer 1

The correct option is D is equal to $MR{v}_0$

Total angular momentum $L=L_1$ (COM frame) + $L_2$ (of COM)

$L_1=Iw=\frac{1}{2}M{R}^{2}w$

Also, $v_o=Rw$ (no slipping)

$L_1=Iw=\frac{1}{2}MR{v}_o$

$L_2=M(\overrightarrow{PC}\times \overrightarrow{v_o})=M{v}_o(3/2R-R)=1/2MR{V}_o$

Thus $L=MR{v}_o$