A $4.00 \mathrm{~kg}$ hollow sphere $\left(\mathrm{I}=2 / 3 \mathrm{MR}^{2}\right)$ is spinning with an angular velocity of $10.0 \mathrm{rad} / \mathrm{s}$. The diameter of the sphere is $20.0 \mathrm{~cm}$. The rotational kinetic energy of the spinning sphere is
A) $1.50 \mathrm{~J}$.
B) $0.90 \mathrm{~J}$.
C) $1.75 \mathrm{~J}$.
D) $1.33 \mathrm{~J}$.
E) $0.75 \mathrm{~J}$.

Question

Quizplus · Accepted Answer

The rotational kinetic energy can be calculated using the formula $K = \frac{1}{2} I \omega^2$, where $I$ is the moment of inertia and $\omega$ is the angular velocity. For a hollow sphere, $I = \frac{2}{3} MR^2$, where $M$ is the mass and $R$ is the radius. Given $M = 4.00 \, \text{kg}$, $\omega = 10.0 \, \text{rad/s}$, and the diameter $= 20.0 \, \text{cm} = 0.2 \, \text{m}$, the radius $R = 0.1 \, \text{m}$. Substituting these values, we get $K = \frac{1}{2} \times \frac{2}{3} \times 4.00 \times (0.1)^2 \times (10.0)^2 = 1.33 \, \text{J}$.

A 4.00 kg4.00 \mathrm{~kg}4.00 kg Hollow Sphere (I=2/3MR2)\left(\mathrm{I}=2 / 3 \mathrm{MR}^{2}\right)(I=2/3MR2) Is Spinning with an Angular Velocity Of

A $4.00 \mathrm{~kg}$ Hollow Sphere $\left(\mathrm{I}=2 / 3 \mathrm{MR}^{2}\right)$ Is Spinning with an Angular Velocity Of