A fly wheel having mass 3 kg and radius 5 m is free to rotate about a horizontal axis. A string having negligible mass is wound around the wheel and the loose end of the string is connected to 3 kg mass. The mass is kept at rest initially and released. Kinetic energy of the wheel when the mass descends by 3 m is _ _ _ _ J . ( g =10 ~m / s ² )

Step-by-step solution

Let the mass of the flywheel be M = 3 kg and its radius be R = 5 m. Let the hanging mass be m = 3 kg and the distance descended be h = 3 m. The flywheel is a disk, so its moment of inertia is I = 1 2 MR^2 . By conservation of energy, the loss in potential energy of the hanging mass equals the gain in kinetic energy of the system. mgh = 1 2 mv^2 + 1 2 I ^2 Since the string does not slip, v = R , so = v R . mgh = 1 2 mv^2 + 1 2 ( 1 2 MR^2)( v R )^2 mgh = 1 2 mv^2 + 1 4 Mv^2 Substituting the values: 3 10 3 = 1 2 (3)v^