At any angular speed,a certain uniform solid sphere of diameter D has half as much rotational kinetic energy as a certain uniform thin-walled hollow sphere of the same diameter when both are spinning about an axis through their centers.If the mass of the solid sphere is M,the mass of the hollow sphere is
A) 3/5 M.
B) 5/3 M.
C) 5/6 M.
D) 6/5 M.
E) 2 M.

Question

Quizplus · Accepted Answer

The rotational kinetic energy of a solid sphere is given by:
$K_{solid} = \frac{2}{5}MR^2\omega^2$
where $M$ is the mass of the sphere, $R$ is the radius (which is half the diameter $D$), and $\omega$ is the angular speed.
The rotational kinetic energy of a thin-walled hollow sphere is given by:
$K_{hollow} = \frac{2}{3}MR^2\omega^2$
Since we are told that the solid sphere has half as much rotational kinetic energy as the hollow sphere, we can write:
$\frac{1}{2}K_{hollow} = K_{solid}$
Substituting the expressions for $K_{solid}$ and $K_{hollow}$ above, we get:
$\frac{1}{2}	imes\frac{2}{3}MR^2\omega^2 = \frac{2}{5}MR^2\omega^2$
Simplifying this equation and canceling out the $R^2\omega^2$ terms, we get:
$M_{hollow} = \frac{3}{2}M_{solid}$
Substituting $R = \frac{D}{2}$ and simplifying, we get:
$M_{hollow} = \boxed{	extbf{(D) }6/5 \; M}$

At Any Angular Speed,a Certain Uniform Solid Sphere of Diameter