A block of mass M is moving with a velocity $v_1$ on a frictionless surface as shown in fig. It passes over to a cylinder of radius R and moment of inertia I which has fixed axis and is initially at rest. When it first makes contact with the cylinder, it slips on the cylinder, but the friction is large enough so that slipping ceases before it losses contact with the cylinder. Finally it goes to the dotted position with velocity $v_2$ compute $v_2$ in terms of $v_1, M, I$ and $R$.
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When the mass M makes first contact with the cylinder, its angular momentum about O. $L_1$ = moment of intertia about O \times angular velocity $=\text{MR}^2\times\alpha=\text{MR}^2\Big(\frac{\text{v}_1}{\text{R}}\Big)=\text{MRv}_1$ When the mass M loses the contact with the cylinder, its final angular momentum, $\text{L}_2=\text{MRv}_2$
$\therefore$ Loss in angular momentum $\Delta\text{L}=\text{L}_1-\text{L}_2=\text{MR}(\text{v}_1-\text{v}_2)\dots(1)$ This provides an angular impulse to the cylinder about O which is given by $\tau\Delta\text{t}.$ Now, Loss in angular momentum of mass = angular impulse to the cylinder $\therefore\text{MR}(\text{v}_1-\text{v}_2)=\tau\Delta\text{t}\dots(2)$$\therefore\tau=\text{I}\alpha=\text{I}\frac{\Delta\omega}{\Delta\text{t}}$
Initially cylinder is at rest and finally it moves with a velocity $v_2$, hence, $\Delta\omega=\frac{\text{v}_2}{\text{R}}$
$\therefore\tau=\text{I}\frac{\text{v}_2}{\text{R}.\Delta}\dots(3)$Substituting the value of $\tau$ from eqn. (3) in eqn. (2), we get
$\text{MR}(\text{v}_1-\text{v}_2)=\frac{\text{I}\text{v}_2.\Delta\text{t}}{\text{R}.\Delta\text{t}}=\text{I}\frac{\text{v}_2}{\text{R}}$
$\text{or }(\text{v}_1-\text{v}_2)=\frac{\text{I}}{\text{MR}^2}\text{v}_2$
$\text{or }\text{v}_1=\text{v}_2\Bigg[1+\Big(\frac{\text{I}}{\text{MR}^2}\Big)\Bigg]$
$\text{v}_2=\frac{\text{v}_1}{\Bigg[1+\Big(\frac{\text{I}}{\text{MR}^2}\Big)\Bigg]}$
$\therefore$ Loss in angular momentum $\Delta\text{L}=\text{L}_1-\text{L}_2=\text{MR}(\text{v}_1-\text{v}_2)\dots(1)$ This provides an angular impulse to the cylinder about O which is given by $\tau\Delta\text{t}.$ Now, Loss in angular momentum of mass = angular impulse to the cylinder $\therefore\text{MR}(\text{v}_1-\text{v}_2)=\tau\Delta\text{t}\dots(2)$$\therefore\tau=\text{I}\alpha=\text{I}\frac{\Delta\omega}{\Delta\text{t}}$
Initially cylinder is at rest and finally it moves with a velocity $v_2$, hence, $\Delta\omega=\frac{\text{v}_2}{\text{R}}$
$\therefore\tau=\text{I}\frac{\text{v}_2}{\text{R}.\Delta}\dots(3)$Substituting the value of $\tau$ from eqn. (3) in eqn. (2), we get
$\text{MR}(\text{v}_1-\text{v}_2)=\frac{\text{I}\text{v}_2.\Delta\text{t}}{\text{R}.\Delta\text{t}}=\text{I}\frac{\text{v}_2}{\text{R}}$
$\text{or }(\text{v}_1-\text{v}_2)=\frac{\text{I}}{\text{MR}^2}\text{v}_2$
$\text{or }\text{v}_1=\text{v}_2\Bigg[1+\Big(\frac{\text{I}}{\text{MR}^2}\Big)\Bigg]$
$\text{v}_2=\frac{\text{v}_1}{\Bigg[1+\Big(\frac{\text{I}}{\text{MR}^2}\Big)\Bigg]}$
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