By Philip Rabinowitz
ISBN-10: 0677142358
ISBN-13: 9780677142357
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Extra resources for Numerical methods for nonlinear algebraic equations
Sample text
There are two components to this displacement, (1) a translation of the whole body through the displacement x ft and (2) a rotation of the body round an axis through the center of gravity which gives it an angular displacement Θ which is equal x/R. For the purpose of our analysis we may legitimately 28 I. RIGID BODY MOTION AROUND A FIXED AXIS separate these two components and consider that the body is first given a pure translation at a velocity v fps through the displacement x ft. During this displacement the kinetic energy of the body is 1 W ± — v2 ft-lb 2 g When the body has been so displaced, we rotate it at an angular velocity of ω rad/sec (which equals vjR) through a displacement of Θ rad, equal to xjR.
After a time At the axis of the shaft will have rotated through an angle ω J i , so the angular momentum vector of the wheel will have turned through the same angle. Its change in direction is shown by the vector Α{ΙΩ) and from Diagram I-2c is given by Δ{ΙΩ) = (ΙΩ)ωΑί 26 I. RIGID BODY MOTION AROUND A FIXED AXIS so limit *ΰβϊ t->o At = (/ v β)ω = L ' and is directed at right angles to (ΖΩ), that is along the X-axis. The torque, L, which this vector represents is round the X-axis +~X DIAGRAM I-2C in the clockwise sense when looking in the positive direction along the X-axis.
600 ft from its axis of rotation at O. 00 sec -1 . 600 (53°8/), with the vertical calculate the acceleration of the wheel around A, the point of contact of the wheel with the surface, the friction which is acting between the wheel face and the surface and the normal reaction of the surface with the wheel. From the geometry of the figure So angle OGA is 90° and 56 II. 600 K Since the wheel does not slip, the horizontal component of the acceleration of A is zero. So this component of the inertial force at A is also zero.
Numerical methods for nonlinear algebraic equations by Philip Rabinowitz
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