{"id":460,"date":"2018-11-06T06:31:34","date_gmt":"2018-11-06T06:31:34","guid":{"rendered":"http:\/\/phyp01.epgpbooks.inflibnet.ac.in\/?post_type=chapter&p=460"},"modified":"2019-04-29T06:09:25","modified_gmt":"2019-04-29T06:09:25","slug":"rotating-frame","status":"publish","type":"chapter","link":"https:\/\/ebooks.inflibnet.ac.in\/phyp01\/chapter\/rotating-frame\/","title":{"rendered":"Rotating Frame"},"content":{"raw":"
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  1. Introduction<\/strong><\/li>\r\n<\/ol>\r\n \r\n

    Newton\u2019s Law of Motion are valid in an inertial frame of reference which can be chosen to be the frame with \u2018fixed stars\u2019. If we are interested in describing the motion of a dynamical system on earth, we will have to take into account the fact that the earth constitutes a non-inertial frame. The earth is in complex motion with respect to the fixed stars. For our purpose it is enough to treat the coordinate system fixed on earth to be undergoing rotations about its axis. We thus have to deal with non-inertial rotating frame of reference. This will also be convenient for the description of the motion of rigid bodies which will be taken later in the subsequent units.<\/p>\r\n\r\n

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    1. Rotating Coordinate System<\/strong><\/li>\r\n<\/ol>\r\n \r\n

      We will consider two sets of coordinate axis. One set of coordinate is fixed in an \u2018inertial frame\u2019 \u00a0and other \u00a0in a frame which is in arbitrary motion with respect to the inertial frame. We will take the non-inertial frame to be a rotating frame. The coordinates in the inertial or in fixed frame will be designated as \u00a0or \u00a0and in the rotating frame as \u00a0or . Let us consider a point P whose position vector is given by the vector \u00a0in the rotating frame. Let \u00a0be the position vector of the origin of the rotating coordinate system with respect to the fixed inertial frame. The position vector of P in the inertial frame is now given by<\/p>\r\n \r\n\r\n\"\"\r\n\r\n\"\"\r\n

      This result is valid for any arbitrary vector , any arbitrary vector being defined as a physical quantity that transforms according to the same rules as the position vector. Thus for any vector<\/p>\r\n \r\n\r\n\"\"\r\n\r\n\"\"\r\n

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      1. Coriolis Force<\/strong><\/li>\r\n<\/ol>\r\n \r\n

        Force on a particle in the rotating frame can be obtained from the expression of Newton\u2019s equation of motion which is valid in the inertial (fixed) frame namely,<\/p>\r\n \r\n\r\n\"\"\r\n\r\n\"\"\r\n

        The \u2018centrifugal\u2019<\/strong> and \u2018coriolis\u2019<\/strong> forces are not forces in the usual sense. Nevertheless the forces are real in the sense that one feels an outward centrifugal force in a merry-go-round. These forces can be dispensed with if we prefer to describe the motion of a dynamical system in the inertial frame only. They arise because we have chosen to describe the motion using Newton\u2019s law in a \u2018non-inertial frame\u2019 in the present case a \u2018rotating frame\u2019. Thus in order to write Newton\u2019s equation in a non-inertial frame we need to add non-inertial terms in the expression of the force along with the usual force<\/p>\r\n \r\n\r\n\"\"\r\n

        The Coriolis force like the centrifugal force arises only in a rotating frame of reference. If for example we want to describe the motion of a particle under gravity on earth, in addition to the force of gravity acting on the particle, we will need to take into account the centrifugal and coriolis force. The coriolis force associated with earth\u2019s rotation is quite weak because the earth rotates about its axis only once in a day. This corresponds to an angular speed . If an object is moving with a speed of 1000 ms-1 , the maximum possible coriolis force on it is of the order of 10-2 ms-2 which is\u00a0much smaller than . Coriolis force may be weak but it has important consequences. On earth it acts to change the direction of a moving body to the right in the Northern Hemisphere and to the left in the Southern Hemisphere.<\/p>\r\n\"\"\r\n

        The centrifugal and coriolis forces are responsible for large scale atmospheric circulation, in the development of storms and in the sea-breeze circulation. It is important to incorporate coriolis force in large distance fights and in long-range ballistics.<\/p>\r\n \r\n

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

        A gunner fired a shot in the Northern hemisphere at a latitude of 450. The gun is fixed at an angle of elevation \u00a0towards north. Show that if the range is 20 Km, the shell falls 44m towards east from the mark and if the gun was fired in the southern hemisphere, the shot would have fallen 88 m towards west.Ans.: Set up the coordinates as shown<\/p>\r\n \r\n\r\n\"\"\r\n

        The z-axis is along the line joining the centre of the earth with the point P on the surface from where the gun is fired. Choose y and x axis as shown such that the gun is fired in the y-z plane. The angular velocity \u00a0has components given as<\/p>\r\n\"\"\r\n

        The velocity of the shot makes an angle \u00a0with the y-axis and is in the y-z plane. Its components are<\/p>\r\n \r\n\r\n\"\"\r\n

        the acceleration along the z-axis is . The effect of centrifugal force which acts along the +ve z direction is to modify the acceleration due to gravity.The equation of motion (17.15) gives<\/p>\r\n \r\n\r\n\"\"\r\n\r\n\"\"\r\n

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        1. The Foucault Pendulum<\/strong><\/li>\r\n<\/ol>\r\n \r\n

          Foucault pendulum is a heavy metal sphere suspended by a long wire. The suspension point of the pendulum is free to rotate in any direction. The effect of coriolis force on the motion of the pendulum is to induce a rotation of the plane of oscillation. We will consider the motion of the pendulum at an latitude . The coordinate axis are chosen such that the z-axis is along the local verticle. The origin is chosen at the point where the pendulum is at rest in equilibrium and x-y axis are chosen in the horizontal plane. We assume the amplitude of oscillations to be small and let T<\/strong> be the tension in the wire which has a length . For small oscillations T<\/strong> can be resolved into components as<\/p>\r\n \r\n\r\n\"\"\r\n

          The tension provides the restoring force proportional to the distance along the x and y directions. The rotational velocity \u00a0at the latitude \u00a0has components given in equation (17.16)<\/p>\r\n\"\"\r\n

          The coriolis force produces small velocity components in the x and y directions and there is no motion in the z-direction, thus<\/p>\r\n\"\"\r\n\r\n\"\"\r\n

          By equating the real and imaginary parts, we can easily obtain the solutions for x and y. We see that over and above the oscillator motion of the pendulum the plane in which the pendulum oscillates also<\/p>\r\n \r\n

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