Question

In this problem we consider rotational motion of a diatomic molecule such as carbon monoxide or nitric oxide. We treat a system of two point masses, mi and m2, rotating about their common center of mass. There are no external forces or torques on the system. We are in the center-of-mass frame, so the CM is at the origin. We treat the case of steady rotation, with w pointing in the z direction, and the particles moving in the ry plane. The distance d between the particles is constant. (a) At time t = 0, m 1 is on the positive y axis and m2 is on the negative y axis. Compute the locations of the particles and each individual particle's contribution to the total angular momentum. From this, compute the total angular momentum. Does it agree with what you expect based on Chapter 8? (b) Still at t = 0, compute the moment of intertia tensor as defined in Section 10.3, and from it compute the angular momentum. All of the calculations for parts (a) and (b) are done at (c) At time t repeat the calculations in parts (a) and (b). Note that the moment of inertia tensor will be different because we are currently using the "space frame" tensor rather than the "body frame" tensor (we will define those terms later in the course)

Answer 1

part (a) d_1 + d_2 = d - 12 m (d_1) + 16 m(d_2)/28 m = 0 = y_cm = 0 d_1/d_2 = 4/3 d_2 (7/3) = d d_2 Rightarrow 3/7 d d_2 Rightarrow 4/7 d I net about Rightarrow (m_1 d_1^2 + m_2 d_2^2) Angular momentum Rightarrow (m_1 (4/7 d)^2 + m_2 (3/7 d)^2) w L vector about cm Rightarrow (16 m_1 + 9 m_2) w d^2/49 L vector about 0 span = [(16 m_1 + 9 m_2) d^2/49 + (m_1 + m_2) a^2] w \r\n [L_x' L_y' L_z'] = [I_x' x' i_y' x' I_z' x' I_x' y' I_y' y' I_z' y' I_x' z I_y' z I_z' z'] [w_x' w_y' w_z'] w_x' = 0 w_y' = 0 w_z' = w_2 = w L_x' = I_x' z' w L_y' = i_y' z w L_z' = I_z' z' w L vector = L_x' i^+ L_y j^+ L_z' k^I_= [I_x' x' I_y' x' I_z' x' I_x' y' I_y' y' I_z' y' I_x' z' I_y' z' I_z' x']