A uniform solid cylinder of mass 2 kg and radius 10 cm is rotated about its central geometric axis with an angular speed of 5 rad s ⁻¹ . If the moment of inertia of the cylinder about a generator (a line on the surface parallel to the central axis) is numerically equal to x 10⁻¹ times its angular momentum about the central axis, then the value of x is

Step-by-step solution

The moment of inertia of a solid cylinder about its central axis is I_ central = 1 2 MR^2 The angular momentum about the central axis is L = I_ central = ( 1 2 MR^2 ) 5 = 5 2 MR^2 Using the parallel axis theorem, the moment of inertia about a generator (a line on the surface parallel to the central axis at a distance R ) is I_ gen = I_ central + MR^2 = 1 2 MR^2 + MR^2 = 3 2 MR^2 According to the given condition, I_ gen = x 10⁻¹ L 3 2 MR^2 = x 10 5 2 MR^2 3 = x 10 5 x = 6 Answer: 6