Rotational Motion

Concept:Rolling without slipping relates the orbital angular velocity $\Omega$ of a small disk’s centre to its spin $\omega'$ via $\Omega (R+r) = \omega' r$.Explanation:Each small disk has radius $r = R/50$. Initially the disks touch each other. Distance between their centres is $2r$. Distance from large disk centre to a small disk centre is $R+r$. In the isosceles triangle formed, chord length $2r = 2(R+r)\sin(\Delta\theta/2)$. Using $\sin x \approx x$: $\Delta\theta/2 = r/(R+r)$. Substitute $R = 50r$: $\Delta\theta = 2r/(51r) = 2/51$ radians. No slip at contact: linear speed of the centre = $\omega' r = \Omega (R+r)$. Hence $\Omega = \omega' r/(R+r) = \omega'/51$. Disk 1 spins at $\omega$: $\Omega_1 = \omega/51$. Disk 2 spins at $2\omega$ in the opposite direction: $\Omega_2 = 2\omega/51$. Relative orbital angular speed $\Omega_{\text{rel}} = \Omega_1 + \Omega_2 = 3\omega/51$. Initial angular separation between centres is $\Delta\theta = 2/51$. To be in contact again, the smallest angular separation must again be $\Delta\theta$. This happens when the separation becomes $2\pi - \Delta\theta$ (after crossing $\pi$). Total relative angular displacement: $\theta_{\text{rel}} = 2\pi - 2\Delta\theta$.