By A. Frank D'Souza

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5. 6 Parallel axes theorem. 5 Cone rolling on horizontal surface. This coordinate system remains parallel to the Oxyz considered previously with origin at 0. The expressions for the angular velocity Q of the cone and for the velocity of center of mass C remain unchanged. Hence, the linear momentum of the cone is still given by f. mvc lmhw. cos rxf coordinate systems. and x 2 Yzz 2 are shown in Fig. 6 with origins at points l and 2 of a rigid body. Let x 2 = x 1 +a, Yz = y 1 + b, and z 2 z 1 +c.

24 Splashdown of satellite. 91), we obtain - \ '-.. _ = G(m 1 + m2) (1 _ e) Ym '\ I 180° and it follows that rm ", ""'- ~ 1/ 0 in 0 = / A = G(m, ,_d m2) (1 + e) e ......................... 89) r. (m-'----,-+-;=-m-2')-"----" and express the ratio rmfr. as a function G(m, + m2) v2 = e = e R- = G(m 1 + m2) r;v; G(m, + m2) r0 (r0 v0 ) 2 (;2 - 1) « Since m 1 m 2 where m 1 and m 2 are the masses of the satellite and earth, respectively, it follows that G(m 1 + m2) ~ Gm 2. Now, on the surface of the earth the weight of m 1 becomes Gm 1 m 2 m, g = rm=~ r0 2 - IX 2 We get a real, positive solution for rm only if IX 2 < 2.

J'J) 82 Dynamics of Rigid Bodies; Newton's Law, Energy, and Momentum Methods Chap. 1 I~ A rigid cone with apex half-angle ex rolls steadily without slip on a horizontal surface so that it precesses about the vertical Z axis at a constant angular rate (J) 0 • The height of the cone is h and its base radius is R. Determine (a) the velocity and acceleration of point Pat the base of the cone shown in Fig. 3, and (b) the velocity and acceleration of C, the mass center of the cone. The coordinate system 0 X YZ is inertial with the Z axis being vertical.

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