A disk of clay is rotating with angular velocity \(\omega .\) A blob of clay is stuck to the outer rim of the disk, and it has a mass \(\frac{1}{10}\) of that of the disk. If the blob detaches and flies off tangent to the outer rim of the disk, what is the angular velocity of the disk after the blob separates? a) \(\frac{5}{6} \omega\) b) \(\frac{10}{11} \omega\) c) \(\omega\) d) \(\frac{11}{10} \omega\) e) \(\frac{6}{5} \omega\)

Let's call the initial mass of the disk \(M_d\), the mass of the blob of clay \(M_b\), and the radius of the disk \(R\). The initial angular momentum of the system (\(L_i\)) is the sum of the angular momentum of the disk and the angular momentum of the blob. The final angular momentum of the system (\(L_f\)) is just the angular momentum of the disk after the blob has separated since the blob is flying off tangent to the outer rim of the disk. We know the mass of the blob is \(\frac{1}{10}\) of that of the disk, so \(M_b = \frac{M_d}{10}\).

To find the initial and final angular momentum of the system, we need to calculate the initial and final moments of inertia. Let's denote the moment of inertia of the disk as \(I_d\) and the moment of inertia of the blob as \(I_b\). Initially, both the disk and the blob rotate around the center of the disk, so the initial moment of inertia of the blob (\(I_{b_i}\)) includes the effect of its rotation around the disk center: \(I_d = \frac{1}{2} M_d R^2\) (Moment of inertia for a disk) \(I_{b_i} = M_b R^2\) (Moment of inertia for a point mass at a distance R) The final moment of inertia for the disk is the same as the initial moment of inertia since its shape and mass haven't changed: \(I_{d_f} = I_d = \frac{1}{2} M_d R^2\)

We can now calculate the initial and final angular momentum of the system: Initial angular momentum: \(L_i = (\frac{1}{2} M_d R^2 + M_b R^2)\omega = (\frac{1}{2} M_d R^2 + \frac{M_d}{10} R^2)\omega\) Final angular momentum: \(L_f = I_{d_f} \omega_f\)

Since no external torques act on the system, the total angular momentum must be conserved. Therefore, \(L_i = L_f\). We can plug in the expressions for the initial and final angular momenta and solve for \(\omega_f\): \((\frac{1}{2} M_d R^2 + \frac{M_d}{10} R^2)\omega = \frac{1}{2} M_d R^2 \omega_f\) Simplifying the equation and solving for \(\omega_f\), we get: \(\omega_f = \frac{5}{6}\omega\)

The angular velocity of the disk after the blob separates is \(\frac{5}{6} \omega\). Therefore, the correct answer is: a) \(\frac{5}{6} \omega\)