{"id":484,"date":"2018-11-06T06:43:04","date_gmt":"2018-11-06T06:43:04","guid":{"rendered":"http:\/\/phyp01.epgpbooks.inflibnet.ac.in\/?post_type=chapter&p=484"},"modified":"2019-04-29T06:10:16","modified_gmt":"2019-04-29T06:10:16","slug":"motion-of-rigid-body-i","status":"publish","type":"chapter","link":"https:\/\/ebooks.inflibnet.ac.in\/phyp01\/chapter\/motion-of-rigid-body-i\/","title":{"rendered":"Motion of Rigid Body I"},"content":{"raw":"
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  1. Introduction<\/strong><\/li>\r\n<\/ol>\r\n \r\n

    A rigid body is defined as a collection of particles which can be taken to have a discrete or continuation distribution such that the inter-particle distances within the rigid body remain fixed and do not vary. It is obvious that the rigid body so defined is an idealized concept just like the definition of a point particle. It is a useful concept and will be used in our treatment of the motion of a rigid body. The transition from the distribution of discrete particles to the volume distribution of continuous distribution of mass is simply achieved by replacing the summation over discrete particles to volume integration over\u00a0 where \u00a0is the mass density and \u00a0is the volume element.<\/p>\r\n \r\n

    To describe the motion of a rigid body we will use two coordinate systems: one an inertial frame\u00a0\u00a0and the second a frame fixed in the body called the \u00a0. A rigid body has six degrees of freedom. These are taken to be the coordinates of the centre of mass (3 in number) which is also taken to coincide with the origin of the body-fixed axis. The orientation of the rigid body is fixed by choosing three suitable independent angles which fix the orientation of the body-fixed coordinates. These angles are normally chosen to be what are known as the Eulerian angles. Thus we have six coordinates to fix the orientation and the centre of mass of a rigid body.<\/p>\r\n\r\n

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    1. Coordinate systems of a moving rigid body<\/strong><\/li>\r\n<\/ol>\r\n

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      Let the origin of the body-fixed coordinate system coincides with the centre of mass of the rigid body. Its position with respect to the inertial coordinate system be given by the vector R .<\/strong> Let P be any point in the rigid body whose position vector with respect to the body-fixed axis be given by \u00a0with components\u00a0 . The position vector of P with respect to the inertial frame is given by the vector .<\/strong><\/p>\r\n \r\n

      We now consider an infinitesimal displacement of the rigid body. It consists of two parts : an infinitesimal translation of the centre of mass represented by the infinitesimal vector.<\/strong>The second part consists of an infinitesimal rotation about the centre of mass resulting in an infinitesimal\u00a0<\/strong>change in the position vector of P which arises due to an infinitesimal rotation by an angle \u00a0and is given by.We thus get<\/p>\r\n \r\n\r\n \r\n\r\n\"\"\r\n

      where\u00a0 \u00a0is the velocity of the centre of mass of the body and is called the \u2018translation\u2019 velocity of the rigid body. \u00a0is the angular velocity of the rotation of the body and its direction is along the axis of rotation. \u00a0is the velocity of the point P in the body with respect to the inertial axis. Thus the velocity of any point in the rigid body can be expressed as a sum of translation velocity and the velocity of rotation.<\/p>\r\n \r\n

      We will now prove an important result: \u2018The angular velocity of rotation of the body-fixed<\/strong> coordinate system is independent of the choice of the body-fixed coordinate system\u2019<\/strong>. To see this,let us shift the origin of the body-fixed axis from O to \u00a0and the vector \u00a0by the vector. Let be the new velocity of the point \u00a0and \u00a0be the new angular velocity of rotation along the new axis of rotation. We would then have<\/p>\r\n \r\n\r\n\"\"\r\n

      Thus if the velocity \u00a0of the C.M. of the rigid body at any given instant is perpendicular to the rotation velocity \u00a0for some choice of the origin, then \u00a0\u00a0and \u00a0are also perpendicular. In that case from (18.2) the velocity of all points in the body are perpendicular to the angular velocity . <\/strong>We can then choose suitable origin \u00a0such that \u00a0can be made equal to \u00a0and thereby from (18.5) \u00a0is zero. Thus the motion of a rigid body at that instant is \u2018pure rotation\u2019<\/strong> about an axis through this new choice\u00a0of the origin. Such axis is called the \u2018instantaneous axis of rotation\u2019.<\/strong>It is however, more convenient to take the origin of the body-fixed frame to be at the centre of mass so that in general, during the motion of a rigid body, both the magnitude and the direction of the angular velocity \u00a0vary.<\/p>\r\n\r\n

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      1. The Inertia Tensor<\/strong><\/li>\r\n<\/ol>\r\n\"\"\r\n\r\n\"\"\r\n\r\n\"\"\r\n\r\n\"\"\r\n\r\n\"\"\r\n

        The diagonal elements \u00a0are called the moment of inertia about the \u00a0and \u00a0axis respectively.<\/p>\r\n \r\n

        If we have a continuous distribution of mass in the rigid body, the sum in (18.12) is replaced by an integral namely<\/p>\r\n \r\n\r\n\"\"\r\n

        where\u00a0 \u00a0is the volume mass density and \u00a0\u00a0is the volume element .The inertia tensor \u00a0being a symmetric tensor of rank-2 can be reduced to a diagonal form by a suitable choice of the axis , \u00a0. Corresponding to these Principal Axis of Inertia the diagonal components of the tensor are called Principal Moments<\/strong> of Inertia denoted by \u00a0and \u00a0. The kinetic energy of rotation of a rigid body about the Principal axis of rotation can be written as<\/p>\r\n \r\n\r\n\"\"\r\n

        If the rigid body has some symmetry, then its centre of mass and principal axis also have the same symmetry. When there are two axis of symmetry, i.e. \u00a0body is called a \u2018symmetric top\u2019<\/strong>.When the body has spherical symmetry i.e. \u00a0the rigid body is called a \u2018spherical top\u2019<\/strong>. An\u00a0\u2018asymmetric top\u2019 <\/strong>has no axis of symmetry and \u00a0and \u00a0are all different.<\/p>\r\n\r\n

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        1. Angular Momentum of a Rigid Body.<\/strong><\/li>\r\n<\/ol>\r\n \r\n

          Angular momentum of a system depends on the point with respect to which it is defined. We saw that for a rigid body, the most appropriate choice of this point is the origin of the rotating body-fixed axis. The origin of the body-fixed axis is in turn is taken to be the centre of mass of the body. Thus from the definition of angular momentum we have<\/p>\r\n \r\n\r\n\"\"\r\n\r\n\"\"\r\n

          And the angular momentum is in the direction of angular velocity and is proportional to it. In general the angular momentum of a rigid body is not in the same direction as its angular velocity. The kinetic energy of rotation \u00a0can be written in terms of angular momentum by using (18.11) and (18.20)<\/p>\r\n\"\"\r\n

          If a rigid body is not subjected to any external force, it moves with a uniform velocity and thus its centre of mass motion can be ignored. For an isolated system not subjected to any forces or torques, the angular momentum is conserved. For a spherical top this implies that its angular velocity is constant and thus the general free rotation of a spherical top is a uniform rotation about a fixed axis in space.<\/p>\r\n \r\n

          Example: <\/strong>Calculate the moment of inertia of a solid right circular cone of mass M, height h and base radius R. Locate the coordinates of the centre of mass.<\/p>\r\n \r\n

          Solution:<\/strong><\/p>\r\n

          \"\"<\/p>\r\n\"\"\r\n\r\n\"\"\r\n\r\n\"\"\r\n\r\nSummary:<\/strong>\r\n