A uniform solid sphere of radius `r = 0.500 m` and mass `m = 15.0 kg` turns counterclockwise about a vertical axis through its centre.Find its vector angular momentum about this axis when its angular speed is `3.00 rad//s`.

Options

A. `(2.50 kg.m^2//s) hat k`
B. `-(4.50 kg.m^2//s) hat k`
C. `-(2.50 kg.m^2//s) hat k`
D. `(4.50 kg.m^2//s) hat k`

Answer

D

Explanation

(d) The moment of inertia of the sphere about an axis through its centre is.
`I = (2)/(5) MR^2 = (2)/(5)(15.0 kg)(0.500 m)^2 = 1.50 kg.m^2`
Therefore, the magnitude of the angular momentum is
`L = I omega = (1.50 kg.m^2)(3.00 rad//s) = 1.50 kg.m^2//s`
Since the sphere rotates counterclockwise about the vertical axis, the angular momentum vector is directed upward in the `+ z` direction. Thus, `vec L = (4.50 kg.m^2//s) hat k`.