Question

1. Find the kinetic energy of the uniform circular cone of height h, base radius R, and mass M. Rotating with the angular velocity ~ around an axis which goes through the center of mass. 2. The same for a uniform ellipsoid of semiaxes a, b, and c. 1 {

Answer 1

Answer:

1) For the uniform circular cone, KE = 0.15MR²Ω²

2) For the ellipsoid, KE = 0.1M(a²+b²)Ω²

Explanation:

1) For a circular cone rotating about the verical axis, the moment of inertia is given by $I_{z} = \frac{3}{10} MR^{2}$

The kinetic energy of an object with angular velocity Ω is given by:

KE = 0.5 I Ω²

KE = o.5 * 0.3 MR²Ω²

KE = 0.15MR²Ω²

2) For a uniform ellipsoid of semi axes a, b, and c rotating in the vertical axis, the moment of inertia is given as $I_{z} = \frac{M}{5} (a^{2} + b^{2} )$

The kinetic energy is given by KE = 0.5 I Ω²

KE = o.5 * 0.2 M(a²+b²)Ω²

KE = 0.1M(a²+b²)Ω²

Answer 2

Answer:

1) $I = \frac{3}{10}\cdot m\cdot r^{2}$, 2) $K = \frac{1}{10}\cdot m \cdot (b^{2}+c^{2})$

Explanation:

1) The kinetic energy due to the rotation is:

$K = \frac{1}{2}\cdot I\cdot \omega^{2}$

Where $I$ and $\omega$ are the moment of inertia and angular speed, respectively. The moment of inertia of the circular cone is:

$I = \frac{3}{10}\cdot m\cdot r^{2}$

The kinetic energy is:

$K = \frac{3}{20}\cdot m\cdot r^{2}\cdot \omega^{2}$

2) The moment of inertia of the ellipsoid is (where a is the major semiaxis):

$I = \frac{1}{5}\cdot m \cdot (b^{2}+c^{2})$

The kinetic energy is:

$K = \frac{1}{10}\cdot m \cdot (b^{2}+c^{2})$