CH6_ANGULAR_MOMENTUM

CH6_ANGULAR_MOMENTUM - ANGULAR MOMENTUM ­ A GENERAL...

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Unformatted text preview: ANGULAR MOMENTUM ­ A GENERAL APPROACH 9.1 Some basic theorems Theorem 1 ((Chasles’ Theorem) The motion of a rigid body can be described by the translation motion of a point P in the body and the rotational about an axis (which may change with time) passing through P. Theorem 2 S1, S2 and S3 are three coordinate systems having the same origin. S1 rotates with angular velocity with respect to S2 and S2 rotates with angular velocity with respect to S3. Then S1 rotates with angular velocity with respect to S3. Proof: Consider points P1, P2 and P3 attached to the coordinate systems S1, S2 and S3 respectively, and they coincide at time t=0. At t=0, the position vector of these points is as relative to the origin. Notice that these three points would have different non‐zero velocities at t=0. Velocity of P1 relative to P2 due to the rotation is: Velocity of P2 relative to P3 due to the rotation is: Therefore the velocity of P1 relative to P3 is: As P1 and P3 are attached to the coordinate systems S1 and S3, S1 rotates with angular velocity with respect to S3. Example: We are now observing the following motion from the laboratory frame S3. A sphere (coordinate S1) is rotating about a stick (coordinate S2) with angular velocity . The stick is rotating with angular velocity about the direction with respect to the laboratory frame. The stick initially points to the direction of direction, where are the unit vector of S3. At time t=0, and Therefore, the angular velocity of the sphere relative the S3 is: As time goes by, the stick (i.e. traces out a cone. ) rotates about y‐axis and moves down, thus 9.2 Inertia tensor A. Rotation about an axis through the origin A rigid body is rotating with the angular velocity with respect to the laboratory frame. Consider a small mass dm of the rigid body at the position of . The angular momentum of this small mass is: (6.1) (6.2) The total angular momentum of the whole rigid body is thus: If and , then, (6.3) (6.4) and thus, Therefore the angular momentum vector can be written in the form of column vector as: L1 ∫ ( y 2 + z 2 ) dm − ∫ xydm ∫ (z 2 + x 2 )dm L2 = − ∫ xydm − ∫ yzdm L3 − ∫ zxdm Ixx Ixy Ixz ω1 = Iyx Iyy Iyz ω 2 I zx Izy Izz ω 3 = Iω ω 1 − ∫ yzdm ω 2 ∫ ( x 2 + y 2 )dmω 3 − ∫ zxdm (6.5) € I is called the inertia tensor, which is dependent on the geometry of the rigid body. I is dependent on the choices of the origin and the x, y, z‐axes. I is a symmetric matrix, i.e. Iab=Iba. There are only six independent entries rather than nine. To derive the KE of the rigid body, consider the small mass dm at position vector which contributes KE of 2: (6.6) (6.7) Therefore the total KE is equal to: Multiplying the expression out and it can be easily proved: (6.8) Example 1: Find the moment of inertia tensor of a cube with mass and length equal to M and L. There two types of integral for calculating the tensor matrix, namely: and Therefore, If the angular velocity of the cube is along the z‐axis, then, Notice that the angular momentum has non‐zero x and y‐components though the angular velocity is in pure z direction. Example 2 Point mass on the x­y plane 0 z=0, x + y = r and ω = 0 ω 2 € 2 2 L1 ∫ ( y 2 + z 2 ) dm ∫ L2 = − € xydm L3 − ∫ zxdm 0 = 0 2 mr ω − ∫ xydm ∫ (z 2 + x 2 ) dm − ∫ yzdm ω 1 − ∫ yzdm ω 2 ∫ ( x 2 + y 2 )dmω 3 − ∫ zxdm € Example 3 Point mass on plane parallel to the xy plane. 0 z=z0 , x + y = r and ω = 0 ω 2 € 2 € 2 Notice that the angular momentum vectors are different for different choices of origin. The angular momentum as found in Example 3 has non‐zero x and y components though the angular velocity is of pure z direction. Moreover, its x and y components are dependent on the particle position. This figure clearly illustrates how the non‐zero x and y angular momentum components are originated. B. General Motion Now consider a rigid body having a general motion consisting of CM translation and rotation about the CM. Notice that unlike the case we have discussed in the section of Inertia Tensor, the small masses constituting the rigid body no longer rotate about the origin(as shown in the figure), and thus we cannot write: as previously in equation (1). Thus we are going to derive the angular momentum and the kinetic energy T for this general motion case. Using Theorem I and taking the CM as the point P, the general motion of the rigid body can be described as the sum of the translation motion of the CM and the rotation about an axis passing through the CM. Let the CM moves with velocity with respect to the fixed lab. frame, and the rigid body rotates with instantaneous angular velocity about the so called CM frame (i.e. with respect to the frame with origin at CM and three axes parallel to the laboratory frame). Let =Position vector of the CM relative to the lab. frame =Position vector of a small mass dm relative to the CM frame =Position vector the small mass dm relative to the laboratory frame Therefore in the CM frame: Referring to the figure, we have: . The velocity of dm relative to the laboratory frame is thus: The angular momentum of the whole rigid body relative to the laboratory frame is by definition equal to: (6.9) However is a constant for all the dm’s because all the them rotates with the same angular velocity with respect to the CM frame. Therefore, (6.10) (6.11) as (6.12) by the definition of the CM. The other cross term:. Therefore finally, where is the angular momentum of the rigid body relative to the CM. The kinetic energy of the rigid body is equal to: However, the cross term (6.13) and thus, (6.14) (6.15) Notice that: Therefore: (6.16) C. Generalized Parallel Axis Theorem Now consider a special case, the CM rotates about the origin with the same angular velocity as that the rigid body rotates about the CM (as shown in the figure). Write as , the CM velocity relative to the origin is thus: . For each of the dm of the rigid body, its velocity relative to the origin is: The total angular momentum of the rigid body relative to the origin is equal to: But: similarly for , we have: (6.17) 9.3 Principal Axes 9.3.1 Diagonalizing the inertia tensor matrix A symmetric matrix M can be diagonalized by the rotational matrix C as: D=C‐1MC where D is the new diagonal matrix. C refers to a rotation about the origin and can be obtained by: where are the three normalized eiganvectors. The three new coordinate axes are along the directions of . This implies for any inertia tensor matrix I which must be symmetric, there exists a rotation of the coordinate system (associated with the rotation matrix C) such that the new I’ is diagonal after the rotation. The directions of these new axes are called principal axes. The diagonal inertia tensor matrix is: and the corresponding I1, I2 and I3 are called the principal moments. If an angular velocity, say , is along the one of the principal axis, then the corresponding angular momentum is given by: or , and similarly, and Similar argument also holds for the other two principal axes. Therefore in general for an angular velocity , its corresponding angular momentum and KE are: and (6.18) (6.19) . We will adopt the principal axis basis for the following discussion unless otherwise specified. Example: For a uniform rectangle, it is very clear that the as drawn and principal axes because and z3=0. axes are Notice that the direction of the principal axes depends on the geometry of the rigid body and thus would move around together with the body if the body rotates. For example considering a body rotating about a principal axis. This implies the other two principal axes rotate around it. Thus the components and appeared in and are measured from the instantaneous principal axes which changes with respect to time. 9.4 Real problem solving I: Motion after an impulsive blow Example: The rigid body with three masses m, m and 2m linked with massless rod is given a kick at B with direction into the page and impulse . What are the velocities of the three masses immediately after the kick? 1. Find the L of the system relative the CM. 2. Find the principal moments, and the angular velocity vector. 3. Find the velocities relative to CM. 4. Add on the CM motion. Find L: Therefore, Remark: Assume is constant as the kick happens very fast. Find the principal moment The x, y, z axes are the principal axes. Find For principal axis basis, That is just after the kick, the rigid body rotate with this relative to the CM. Find the velocities of the three masses relative to CM Find the velocities of the three masses relative to the lab. frame and Thus, 9.5 Real problem solving II: Frequency of motion due to torque A uniform stick with length l and mass m is pivoted at one end and swings around the vertical axis. Assume that stick makes an angle θ to the vertical, fine the frequency of oscillation. (1) Find the principal moment. The principal axis is along the stick, and any other two axes perpendicular to the stick. x‐ and y‐ axes are shown in figure and z‐axis pointing out of the paper. Relative to the pivot, (2) Find relative to the pivot (3) Find The vector traces out a cone surface as the stick rotates around. The tip of undergoes a circle motion with radius of . Thus, Notice that (refer to Section 1.2.6) is perpendicular to y‐axis and , and points in the paper. (4) Find the external torque and equate to 9.6 Euler Equation If we express the angular momentum in terms of the principal axis frame (i.e. the body frame), would have the neat form: but the body frame would change with time if observed from the fixed lab. frame. The change of is thus determined by two factors, namely the change of the body frame, and the influence of the external torque. By equation (3.6): , we can write: Expanding the right side, Euler Equation. 9.6.1 Example: Free symmetric top Considering a symmetric top with I1=I2=I. x1 and x2 form a plane perpendcular to x 3 . View from the body frame Euler equations: The last equation implies ω3 is a constant, take The first two equations thus become: Differentiating the first equation and use the second to eliminate , we have: Solving thus yields: The angular velocity spans out a cone about x3 with frequency Ω, which is dependent on ω3 and the body geometry. The radius of the cone (i.e. dependent on A) is determined by the initial condition. The angular momentum is: Case 1: Ω>0 (or I3>I) This implies: Thus lies above as shown in Fig.?? This is called oblate top. One of the example is a coin. Case 2: Ω<0 (or I3<I) This implies: Thus lies below as shown in Fig.?? This is called prolate top. View from the fixed lab. frame In terms of the changing principal axes : Eliminating thus gives: This implies the three vectors and are co‐planar. We also notice that as there is no external torque, is constantly fixed in the lab. frame. Thus implying and precess around . Moreover as is fixed in the body frame , we notice: This simply implies rotates about with frequency of ...
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This note was uploaded on 08/17/2011 for the course PHYS 230 taught by Professor Harris during the Winter '07 term at McGill.

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