Question

A solid sphere of mass $M$ and radius $R$ is rolling without slipping as shown in figure. Find out the magnitude of total angular momentum of the sphere about point of contact $\left(P\right)$.

• A. $L_{\text{Total}}=\frac{3}{5}M{V}_{0}R$
• B. $L_{\text{Total}}=\frac{12}{5}M{V}_{0}R$
• C. $L_{\text{Total}}=\frac{2}{5}M{V}_{0}R$
• D. $L_{\text{Total}}=\frac{7}{5}M{V}_{0}R$

Answer 1

The correct option is D $L_{\text{Total}}=\frac{7}{5}M{V}_{0}R$

For pure rolling of sphere,
$V_0=R\omega ...\left(i\right)$
$\Rightarrow$Total angular momentum of sphere about point $P$ is given as vector sum of angular momentum about $P$ due to translation of its $COM$ $\&$ angular momentum of sphere about its $COM$ due to rotation.
${\overrightarrow{L}}_{\text{Total}}={\overrightarrow{L}}_{\text{Trans}}+{\overrightarrow{L}}_{\text{Rot}}...\left(i\right)$
$\because$The body is rotating in clockwise direction and moment of linear momentum has clockwise sense of rotation.


Hence, both ${\overrightarrow{L}}_{\text{Trans}}$ $\&$ ${\overrightarrow{L}}_{\text{Rot}}$ will add up.
$\Rightarrow$Angular momentum due to translation is given by:
$L_{\text{Trans}}=mv{r}_{\perp }$
where $v$ is linear velocity of centre of mass, $m$ is mass of body
$r_{\perp }=R$ is the perpendicular distance between linear momentum $M{V}_0$ and point $\left(P\right)$ about which angular momentum is calculated.
$\Rightarrow L_{\text{Trans}}=M{V}_{0}R...\left(ii\right)$
Angular momentum due to rotation is given by,
$L_{\text{Rot}}=I_{\text{CM}}\omega$
$I_{\text{CM}}=\frac{2}{5}M{R}^2$ is the moment of inertia of sphere about axis passing through its centre.
$\Rightarrow L_{\text{Rot}}=\frac{2}{5}M{R}^2\omega$
$\therefore L_{\text{Rot}}=\frac{2}{5}M{R}^2\times \frac{V_0}{R}=\frac{2}{5}M{V}_{0}R...\left(iii\right)$
Hence from Eq $\left(i\right),\left(ii\right)\&\left(iii\right)$ magnitude of total angular momentum of sphere about point $P$ is given as:
$L_{\text{Total}}=L_{\text{Trans}}+L_{\text{Rot}}$
$\Rightarrow L_{\text{Total}}=M{V}_{0}R+\frac{2}{5}M{V}_{0}R$
$\therefore L_{\text{Total}}=\frac{7}{5}M{V}_{0}R$