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--- 183_notes:rot_ke [2022/10/31 14:09] – valen176
+++ 183_notes:rot_ke [2023/11/07 16:42] (current) – hallstein
@@ Line 3: / Line 3: @@
 ===== Rotational Kinetic Energy =====
-Earlier, you read about [[183_notes:energy_sep|how to separate the different forms of kinetic energy]] (translation, vibrational, and rotational). **In this set of notes, you will read about the kinetic energy that is due to rotation about the center of mass. In these notes, you will also be introduced to the moment of an inertia -- a conceptual tool that arises because some things are more difficult to rotate than others.**((The analogy here is to the mass of an object, which describes the "ease" with which an object can be accelerated.))
+Earlier, you read about [[183_notes:energy_sep|how to separate the different forms of kinetic energy]] (translation, vibrational, and rotational). **In this set of notes, you will read about the kinetic energy that is due to rotation about the center of mass. In these notes, you will also be introduced to the moment of inertia -- a conceptual tool that arises because some things are more difficult to rotate than others.**((The analogy here is to the mass of an object, which describes the "ease" with which an object can be accelerated.))
 ==== Atoms in Rotating Objects Can Move with Different Speeds ====
@@ Line 26: / Line 26: @@
 ==== The Rotation of Rigid Objects ====
-{{ 183_notes:week10_rotational1.png?400}}
+{{ 183_notes:week10_rotational1a.png?400}}
 The merry-go-round is an example that demonstrates that you need to keep track of how far objects are from the center of mass when they are rotating. But to determine the kinetic energy of the merry-go-round can be tough because we have to consider how each atom contributes to the kinetic energy. You will read how to do that in a bit, but for now consider the system in the figure to the right that rotates at a constant angular speed,  $\omega$ .
-The kinetic energy of this system is the sum of individual kinetic energies of the four individual particles.
+The kinetic energy of this system is the sum of the individual kinetic energies of the four individual particles.
-$$K = \dfrac{1}{2}{m_1}\,{v_1^2} + \dfrac{1}{2}{m_2}\,{v_2^2} + \dfrac{1}{2}{m_3}\,{v_3^2} + \dfrac{1}{2}{m_4}\,{v_4^2}$$
+$$K = \dfrac{1}{2}{m_1}\,{v_1^2} + \dfrac{1}{2}{m_2}\,{v_2^2} + \dfrac{1}{2}{m_3}\,{v_3^2} + ...$$
 Each of these velocities can be described in terms of the distance to the particles and the angular speed of the whole apparatus (e.g.,  $v_1 = r_{\perp 1}\omega$ ).
-$$K = \dfrac{1}{2}{m_1}\,{(r_{\perp 1}\omega)^2} + \dfrac{1}{2}{m_2}\,{(r_{\perp 2}\omega)^2} + \dfrac{1}{2}{m_3}\,{(r_{\perp 3}\omega)^2} + \dfrac{1}{2}{m_4}\,{(r_{\perp 4}\omega)^2}$$
+$$K = \dfrac{1}{2}{m_1}\,{(r_{\perp 1}\omega)^2} + \dfrac{1}{2}{m_2}\,{(r_{\perp 2}\omega)^2} + \dfrac{1}{2}{m_3}\,{(r_{\perp 3}\omega)^2} + ...$$
 You can group the terms and find that there's a relationship between the total kinetic energy and the angular speed.
-$$K = \dfrac{1}{2}\left({m_1}\,r^2_{\perp 1}+{m_2}\,r^2_{\perp 2}+{m_3}\,r^2_{\perp 3}+{m_4}\,r^2_{\perp 4}\right)\omega^2$$
+$$K = \dfrac{1}{2}\left({m_1}\,r^2_{\perp 1}+{m_2}\,r^2_{\perp 2}+{m_3}\,r^2_{\perp 3}+...\right)\omega^2$$
 The use of the symbol  $\perp$  here is to remind you that the distance measured from the object to the rotation axis is the perpendicular distance to the axis.
@@ Line 107: / Line 107: @@
 ==== Examples ====
-[{{ 183_notes:pp_vs_real.003.png?300|A circular disk rotates at a constant rate.}}]
+[{{ 183_notes:semi_hollow_sphere.png?220|A Semi-Hollow Sphere}}]
+Suppose that you want to find the moment of inertia of a semi-hollow sphere (shown to the right) with outer radius R, inner radius r, and uniform density d rotating about its center. There is no obvious equation for this but you do have an equation for the moment of inertia of a solid sphere, which is $I = \dfrac{2}{5} mR^2 $. To find the moment of inertia of the semi-hollow sphere ($ I_{\text{semi-hollow}} $) you can find the moment of inertia of the outer sphere as if were is **not** hollow ($ I_R $), then subtract the moment of inertia of the inner sphere ($ I_r$) as if it is not really empty space, that is:
+ $I_{\text{semi-hollow}} = I_R - I_r = \frac{2}{5}m_R R^2 - \frac{2}{5}m_r r^2$
+However, you don't know the masses of the outer and inner spheres ( $m_R$  and  $m_r$ ), so you will need to find them. This can be done by first finding their volumes ( $V_R$  and  $V_r$ ) with the volume of a sphere equation:
+ $V_R = \frac{4}{3}\pi R^3, V_r = \frac{4}{3}\pi r^3$
+Then multiplying these volumes by the density of the sphere material:
+ $m_R = \frac{4}{3}\pi R^3d, m_r = \frac{4}{3}\pi r^3d$
+Plugging these back into the moment of inertia equation gives:
+ $I_{\text{semi-hollow}}  = \frac{2}{5}(\frac{4}{3}\pi R^3d)R^2 - \frac{2}{5}(\frac{4}{3}\pi r^3d)r^2 = \frac{8}{15}\pi d(R^5 - r^5)$
-Other examples:
+More examples:
   * [[:183_notes:examples:The Moment of Inertia of a Diatomic Molecule]]
   * [[:183_notes:examples:The Moment of Inertia of a Bicycle Wheel]]