Question

Consider a massless rod connected to two cylinders which are free to rotate about the rod (like tyres of roller skates) , each of mass m and radius R. Length of rod is d. One cylinder has rough curved surface and is given angular velocity omega not and other cylinder has smooth curved surface and is at rest. The system is kept on a horizontal frictionless surface such that, curved surface of cylinders is in contact with the horizontal surface. Find the angular velocity omega with which the rod will rotate finally

Answer 1

The final angular velocity of the rod is $\omega = \frac{2R}{d} \omega_0$.

Answer 2

The friction force on the rough cylinder (Cylinder 1) causes it to roll without slipping, meaning its linear velocity of the center of mass $v_{cm1}$ becomes equal to its angular velocity $\omega_1$ times the radius $R$. Since the total linear momentum of the system is conserved (initially zero), the centers of the two cylinders move with equal and opposite velocities: $v_{cm1} = -v_{cm2}$. If the rod rotates with angular velocity $\Omega$, then $v_{cm1} = \Omega d/2$. The condition for rolling without slipping for Cylinder 1 is $v_{cm1} = \omega_1 R$. Thus, $\Omega d/2 = \omega_1 R$. The friction force $f_1$ acting on Cylinder 1 provides an impulse that changes its linear and angular momentum. The impulse on the center of mass is $\int f_1 dt = m v_{cm1,f}$ and the impulse on angular momentum is $R \int f_1 dt = I_1 (\omega_{1,f} - \omega_0)$. Substituting $v_{cm1,f} = \Omega d/2$ and $\omega_{1,f} = \frac{\Omega d}{2R}$, and assuming $I_1 = \frac{1}{2}mR^2$, we solve for $\Omega$. The final angular velocity of the rod is $|\Omega| = \frac{2R}{d} \omega_0$.