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MIT16_07F09_Lec22

# MIT16_07F09_Lec22 - J Peraire S Widnall 16.07 Dynamics Fall...

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J. Peraire, S. Widnall 16.07 Dynamics Fall 2008 Version 1.0 Lecture L22 - 2D Rigid Body Dynamics: Work and Energy In this lecture, we will revisit the principle of work and energy introduced in lecture L11-13 for particle dynamics, and extend it to 2D rigid body dynamics. Kinetic Energy for a 2D Rigid Body We start by recalling the kinetic energy expression for a system of particles derived in lecture L11, 1 2 1 2 T = 2 mv G + 2 m i r ˙ i , n i =1 where n is the total number of particles, m i denotes the mass of particle i , and r i is the position vector of particle i with respect to the center of mass, G . Also, m = i n =1 m i is the total mass of the system, and v G is the velocity of the center of mass. The above expression states that the kinetic energy of a system of particles equals the kinetic energy of a particle of mass m moving with the velocity of the center of mass, plus the kinetic energy due to the motion of the particles relative to the center of mass, G . For a 2D rigid body, the velocity of all particles relative to the center of mass is a pure rotation. Thus, we can write r ˙ i = ω × r i . Therefore, we have n n n 1 1 1 2 m i r ˙ i 2 = 2 m i ( ω × r i ) · ( ω × r i ) = 2 m i r i 2 ω 2 , i =1 i =1 i =1 where we have used the fact that ω and r i are perpendicular. The term i n =1 m i r i 2 is easily recognized as the moment of inertia, I G , about the center of mass, G . Therefore, for a 2D rigid body, the kinetic energy is simply, T = 1 mv 2 + 1 I G ω 2 . (1) 2 G 2 1

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When the body is rotating about a fixed point O , we can write I 2 O = I G + mr G and 1 2 1 2 1 T = 2 mv G + 2 ( I O mr G ) ω 2 = 2 I O ω 2 , since v G = ωr G . The above expression is also applicable in the more general case when there is no fixed point in the motion, provided that O is replaced by the instantaneous center of rotation. Thus, in general, 1 T = I C ω 2 . 2 We shall see that, when the instantaneous center of rotation is known, the use of the above expression does simplify the algebra considerably. Example Rolling Cylinder Consider the cylinder rolling without slipping on a ﬂat plane. The friction between the plane and the cylinder insure the no-slip condition, but the friction force does no work. The no-slip condition requires ω = V G /R 0 ). We take the cylinder to have its center of mass at the center of the cylinder but we allow the mass distribution to be non-unform by allowing a radius of gyration k G . For a uniform cylinder k G = R 0 / 2 The total kinetic energy, from equation (1) is given by 1 1 1 1 G V 2 m ( R 2 I C ω 2 T = mV 2 + mk 2 G /R 2 = + k 2 ) ω 2 = (2) 2 G 2 0 2 0 G 2 where the parallel axis theorem has been used to relate the moment of inertia about the contact point C to the moment of inertia about the center of mass.
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