※ 引述《VanderPol (Critical Point)》之铭言： : 题目如下: : The spring-mounted collar oscillates on the shaft according to x=0.04sinπt : ,where x is in meters and t is in seconds.Simultaneously the frame rotates : about the bearing at O with an angular velocity ω=2sin(πt/2) rad/s. : Determine the acceleration of the center C of the collar (a)when t=3s and : (b)when t=0.5s. y : | : | : │ ┌───┐ │ : ├─┤ .C ├─┤------------x : │ └───┘ │ | : │ │ | : ＼ ／ | 0.2m : ＼ ／ | : ＼ ／ | : ╭ ╲／------------------- : ╰>ω O Rotating Frames and the Coriolis Effect The procedure of differentiating a vector function by differentiating its components is valid only if the basis vectors themselves do not depend on the variable of differentiation. In some situations in mechanics this is not the case. For instance, in modelling large scale weather phenomena the analysis is affected by the fact that a coordinate system fixed with respect to the earth is, in fact, rotating (along with the earth) relative to directions fixed in space. In order to understand the effect that the rotation of the coordinate system has on representations of velocity and acceleration, let us consider two Cartesian coordinate frames (that is, systems of axes), sharing a common origin, one "fixed" in space and the other rotating with angular velocity Ω about an axis through that origin. Let I, I, K be the standard basis vectors of the fixed coordinate frame, and let i, j, and k be the basis vectors of the rotating frame. The latter three are constant vectors in their own frame but are functions of time in the fixed frame. We showed above that the velocity of a point with position r rotating about an axis with angular velocity Ω is Ω×r. There fore di/dt = Ω×i dj/dt = Ω×j dk/dt = Ω×k Any vector function can be expressed in terms of either basis. If r(t) is the position vector of a moving particle it can be expressed in the fixed frame as r = X(t)I + Y(t)J + Z(t)K or in the rotating frame as r = x(t)i + y(t)j + z(t)k The velocity of the moving particle can be expressed in the fixed frame, F, as V_F = (dX/dt)I + (dY/dt)J + (dZ/dt)K Similarly, the velocity of the particle, measured with respect to the rotating frame, R, is V_R = (dx/dt)i + (dy/dt)j + (dz/dt)k These vectors are, however, not equal. (For instance, a point at rest with respect to the rotating frame would be seen as moving with respect to the fixed frame.) If we want to express the fixed-frame velocity of the particle, V_F, in terms of the rotating basis, we must also account for the rate of change of the basis vectors in the rotating frame; using three applications of the Product Rule we have V_F = (dx/dt)i + x(di/dt) + (dy/dt)j + y(dj/dt) + (dz/dt)k + z(dk/dt) = V_R + xΩ×i + yΩ×j + zΩ×k = V_R + Ω×r This formula shows that differentiation with respect to time has different effects in the two coordinate systems. If we denote by (d/dt)_F and (d/dt)_R the operators of time differentiation in the two frames then the formula above can be written (d/dt)_F r(t) = ((d/dt)_R + Ω×) r(t) or, in terms of operators, (d/dt)_F = ((d/dt)_R + Ω×) These operators can be applied to any vector function, and in particular to the function V_F = V_R + Ω×r. Since (d/dt)_F V_F is the acceleration A_F of the particle in the fixed frame, and (d/dt)_R V_R is the acceleration A_R in the rotating frame, we calculate, assuming that the angular velocity Ω of the rotating frame is constant, A_F = (d/dt)_F (V_R + Ω×r) = ((d/dt)_R + Ω×)(V_R + Ω×r) = A_R + 2Ω×V_R + Ω×(Ω×r) The term 2Ω×V_R is called the Coriolis acceleration and the term Ω×(Ω×r) is called the centripetal acceleration. In order to interpret these accelerations, suppose that the origin is at the centre of the earth, the fixed frame is fixed with respect to the stars, and the rotating frame is attached to the rotating earth. Suppose that the particle has mass m and is acted upon by an external force F. If we view the moving particle from a vantage point fixed in space (so that we see the fixed frame as unmoving) then, in accordance with Newton's second law, we observe that F = m A_F. If, instead, we are sitting on the earth, and regarding the rotating frame as stationary, we would still want to regard A_R as the force per unit mass acting on the particle. Thus we would write A_R = (F/m) - 2Ω×V_R - Ω×(Ω×r) and would interpret the Ω terms as other forces per unit mass acting on the particle. The term -Ω×(Ω×r) is a vector pointing directly away from the axis of rotation of the earth and we would interpret it as a centrifugal force. (If the earth were spinning fast enough, we would expect objects on the surface to be flung into space.) The term -2Ω×V_R is at right angles to both the velocity and the polar axis of the earth. We would call this a Coriolis force. The centrifugal and Coriolis forces are not "real" forces acting on the particle. They are fictitious forces which compensate for the fact that we are measuring acceleration with respect to a frame which we are regarding as fixed although it is really rotating and hence accelerating. --

※ 发信站: 批踢踢实业坊(ptt.cc) ◆ From: 114.44.124.66