- Before the disc is brought in contact with table, it is only rotating about its horizontal axis. So, its centre of mass is at rest i.e. vcm = 0.
- (M) When rim is placed in contact with the table then its linear velocity at any point on the rim of disc will reduce due to kinetic friction.
- If rotating disc is placed in contact with the table, its centre of mass acquires some velocity (which was zero before contact) due to kinetic friction. So linear velocity of CM will increase.
- Kinetic friction.
- Rolling will begin when vCM = R(x).
- Acceleration produced in centre of mass due to friction,
$\text{a}_{\text{cm}}=\frac{\text{F}}{\text{m}}=\frac{\text{u}_\text{k}\text{mg}}{\text{m}}=\mu_\text{k}\text{g}$
Angular acceleration produced by the torque dur to friction.
$\alpha=\frac{\tau}{\text{I}}=\frac{\mu_\text{k}\text{mgR}}{\text{I}}$
$\therefore\ \nu_\text{cm}=\upsilon_\text{cm}+\text{a}_\text{cm}\text{t}$
$\Rightarrow\ \nu_\text{cm}=\mu_\text{k}\text{gt}$
and $\omega=\omega_0+\alpha\text{t}$
$\Rightarrow\omega=\omega_0-\frac{\mu_\text{k}\text{mgR}}{\text{I}}\text{t}$
For rolling without slipping,
$\frac{\nu_\text{cm}}{\text{R}}=\omega_0-\frac{\mu_\text{k}\text{mgR}}{\text{I}}\text{t}$
$\Rightarrow\ \frac{\mu_\text{k}\text{gt}}{\text{R}}=\omega_{\circ}-\frac{\mu_\text{k}\text{mgR}_\text{t}}{\text{I}}$
$\Rightarrow\text{t}=\frac{\text{R}\omega_0}{\mu_\text{k}\text{g}\big(1+\frac{\text{mR}^2}{\text{I}}\big)}$
