Answer:
1) [tex]I = \frac{3}{10}\cdot m\cdot r^{2}[/tex], 2) [tex]K = \frac{1}{10}\cdot m \cdot (b^{2}+c^{2})[/tex]
Explanation:
1) The kinetic energy due to the rotation is:
[tex]K = \frac{1}{2}\cdot I\cdot \omega^{2}[/tex]
Where [tex]I[/tex] and [tex]\omega[/tex] are the moment of inertia and angular speed, respectively. The moment of inertia of the circular cone is:
[tex]I = \frac{3}{10}\cdot m\cdot r^{2}[/tex]
The kinetic energy is:
[tex]K = \frac{3}{20}\cdot m\cdot r^{2}\cdot \omega^{2}[/tex]
2) The moment of inertia of the ellipsoid is (where a is the major semiaxis):
[tex]I = \frac{1}{5}\cdot m \cdot (b^{2}+c^{2})[/tex]
The kinetic energy is:
[tex]K = \frac{1}{10}\cdot m \cdot (b^{2}+c^{2})[/tex]
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 [tex]I_{z} = \frac{3}{10} MR^{2}[/tex]
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 [tex]I_{z} = \frac{M}{5} (a^{2} + b^{2} )[/tex]
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
