A solid sphere of mass 500 g radius 5 cm is rotated about one of its diameter with angular speed of 10 rad s - 1 . If the moment of inertia of the sphere about its tangent is x × 10 - 2 times its angular momentum about the diameter. Then the value of x will be

Step-by-step solution

The angular momentum about the diameter is L diameter  = 2 5 M R 2 ω ; The moment of inertia of a sphere is I C M = 2 5 M R 2 . The moment of inertia about the tangent is I tangent  = I C M + M R 2 = 2 5 M R 2 + M R 2 = 7 5 M R 2 . It is given that ω = 10   rad   s - 1 Using the given relation between the moment of inertia and the angular momentum, 7 5 M R 2 = x × 10 - 2 × 2 5 M R 2 ω ⇒ x = 3 . 5 × 10 2 ω = 3 . 5 × 10 = 35