Analysis of Instantaneous Center of Zero Acceleration of Rigid Body in Planar Motion

Transcription

1 Moern pplie Science pril, 009 nalysis of Instantaneous Center of Zero cceleration of Rigi Boy in Planar Motion Haibin Sun (Corresponing author) & Tingting Liu Department of Physics an Electronics Taishan University Shanong 7101, China bstract When the angular velocity vector of rigi boy in the plane motion unequal to zero, we can certainly fin a point which instantaneous acceleration equal to zero at that particular instant. The point is name instantaneous center of zero acceleration. We prove the existence of the point an obtaine the location of the point. We coul solve mechanical problem quickly using instantaneous center of acceleration as base point. Keywors: Instantaneous center of zero acceleration, Rigi boy, Planar motion 1. Introuction Let us analyze a rigi boy which is in planar motion. It is clear that if we coul fin a point, C, in the rigi boy for which the instantaneous velocity is zero, then the velocity of the boy at that particular instant woul consist only of a pure rotation of the boy about that point (no translation)(frey R. Zypman,005). If we choose the point C which is momentarily stationary at the instant consiere as a base point, the velocity equation for an arbitrary point in the boy simplifies to v =ω r, (1) where ω is the boy angular velocity (J. F. LI,001). If we know the angular velocity, an the raius vector of point with respect to C, then we coul etermine the velocity of point. Point C can be taken as the center of rotation only at the given instant, an is therefore known as the instantaneous center of zero velocity. If we can etermine the instantaneous center of zero acceleration, we can solve some mechanical problems efficiently.. Deuction of instantaneous center of zero acceleration The acceleration of an arbitrary point P in the rigi boy is (SH M CH, 001) a=a + ω (r r0 ) ω (r r 0 ). () Here, is a base point, r is the raius vector of the point consiere with respect to the fixe point O, r 0 is the raius vector of the point with respect to the point O, r is the raius vector of the point with respect to the point, an r =r r 0 ; ω is the angular velocity, ω α = is the angular acceleration. The irections of the angular velocity an angular acceleration inicate the axis of rotation. Consier the base point an the point P having coorinates (ξ o,η o ) an (ξ,η) in a fixe Cartesian coorinate system, o-ξη. In Cartesian coorinate system, - xy, which is a boy fixe reference system, the point P having coorinates (x, y). Take Z axis to be the fixe irection perpenicular to the planar motion. The position of point an P are illustrate in Fig Existence proof When the rigi boy moves in plane with angular velocity, ω, which is unequal to zero, we choose a point, C, in the boy for which the instantaneous acceleration is zero. We efine this point as the instantaneous center of zero acceleration. From Eq. (), we have 0=a + ω (rc r 0 ) ω (r c r 0 ), (3) 191

2 Vol. 3, No. 4 Moern pplie Science where, r c = ξ c i+ η c j is the raius vector of the instantaneous center of zero acceleration with respect to the origin O. Taking the scalar prouct of Eq. () with ω, we obtain 0=ω a +ω [ ω (rc r 0 )] ω [ ω (r c r 0 )]. (4) The irection of angular velocity vector, ω, is parallel to the irection of angular acceleration vector, α = ω. So ω is orthogonal to ω (rc r 0 ). Hence, ω [ ω (rc r 0 )]=0. (5) In fact, the angular velocity, ω, is perpenicular to the plane that the boy is moving, an the relative raius vector, r c r 0, also lies in this plane. So the vector angular velocity, ω, is perpenicular to the vector r c r 0. Therefore, we have ω [ ω (r c r 0 )]=0. (6) Substituting Eqs. (5) an (6) into Eq. (4), we obtain ω a =0. (7) In the case, the acceleration of any point in the rigi boy, a, must be orthogonal to the angular velocity, ω. Surely Eq. (7) is an ientity. In conclusion, when the rigi boy moves in plane with angular velocity, ω, that is unequal to zero, we can certainly fin the point, C, in the boy for which the instantaneous acceleration is zero, i.e., the instantaneous center of zero acceleration must exist.. Position of the instantaneous center of zero acceleration Taking the vector prouct of Eq. () with ω, we obtain ω (rc r 0 )] ω [ ω (r c r 0 )]. (8) itionally, 0=ω a +ω [ ω [ ω (rc r 0 )]=α [(r c r 0 ) ω] (r c r 0 )(ω α )= (r c r 0 )(ω α ). (9) Here, we have use the vector ientity: (B C) = ( C)B ( B)C ( K. F. Riley, 00). Substituting Eq. (9) into Eq. (8), we have 0=ω a (r c r 0 )(ω α ) (r c r 0 )(ω α ). (10) From Eq. (10), we can obtain the raius vector of the instantaneous center of zero acceleration. The calculation of r c, in this case, requires knowing r 0, a, ω an α. In general, the angular acceleration vector,α, is unknown to us. Therefore, the location of the instantaneous center of zero acceleration is ifficult to locate. In the fixe Cartesian coorinate system o-ξη, Eq. (10) can be written as ω a αa ξ η ξc = ξ0 + 4 αa ξ + ω a η ηc = η0 + 4 α + ω In the moving Cartesian coorinate system -xy, Eq. (10) can be written as, (11) 19

3 Moern pplie Science pril, 009 ω a αa x y xc = 4 αax + ω ay yc = 4 α + ω, (1) It shoul be note that the location of the instantaneous center of zero acceleration will change in time. The path escribe by the instantaneous center is calle the space centroe, an the locus of the positions of the instantaneous centers on the boy is calle the boy centroe. 3. pplication of the instantaneous center of zero acceleration 3.1 cceleration of any point in the rigi boy If we efine the instantaneous center of zero acceleration, C, as the base point, the acceleration of any point, D, in the boy can be obtaine from Eq. (), Here, ω = ω (r r c ) ω (r r c )= [ ω ( r r )] c. (13) ω (r rc ), is the tangential component of the acceleration with respect to the instantaneous center, an, (r r c ), is the normal component of the acceleration with respect to the instantaneous center. The above equation is analogous to the equation expressing the acceleration to fixe axis rotation. So we can say that the acceleration of the boy at that particular instant woul consist only of a pure rotation of the boy about the instantaneous center of acceleration. We choose the instantaneous center of acceleration as the origin of the Cartesian coorinate, in this coorinate system, Eq. (13) has the form = ω r ω r= ( ω r), (14) where r = CD uuur =xi+yj is the raius vector of point D with respect to the instantaneous center, C (cf. Fig. ). ssume that the angular velocity, ω, an the angular acceleration, ω α =, have the same irection. In component form, Eq. (14) can be written as = adx ω x α y ady = αx ω y, (15) 3. Location of the instantaneous center of acceleration t a particular instant, the angular velocity, ω, the angular acceleration, α = ω, an the acceleration of point D, = x i+ y j, all have certain values. Solving Eqs. (15), we have ω a αa x = αa + ω a y = α + ω Dx Dy 4 Dx Dy 4, (16) From Eqs. (16), we can obtain the istance between point D an instant center C: r = CD = x + y = Here, a = a + a, is the magnitue of acceleration. D Dx Dy 4 α + ω. (17) 193

4 Vol. 3, No. 4 The angle, ϕ, between the irection of acceleration vector an the irection of raius vector r is Moern pplie Science ϕ = arctan α. (18) ω Here, if α >0, ϕ is a positive acute angle, i.e., counterclockwise to point D; otherwise, ϕ clockwise to point D. ccoring to Eqs. (17) an (18), if we know the values of ω, α an a D, we can calculate the angle,ϕ, an the magnitue of CD uuur. Then we can raw the location of the instantaneous center of zero acceleration, C (Ch H Ch, 1999). See Fig Theorem of relative angular momentum about the instantaneous center The equation of relative angular momentum of a rigi boy which is in planar motion is (H. E. Williams, 000) Here, point C about an L M = + mrc a. (19) M is the net moment about, L is the relative angular momentum about, a is the acceleration of point. r C is the raius vector of When we efine the instantaneous center of zero acceleration, C, as the base point, the theorem of relative angular momentum with respect to the instantaneous center can be expresse as M c = L = Iα (0) Here, M c is the net moment about point C of all external forces acting on the boy, L is the relative angular momentum about C of the boy, I is the moment of inertia about C, α is the angular acceleration of the boy. 4. Conclusion On the basis of vector algebra, we have prove the existence of the instantaneous center of zero acceleration, an calculate the location of the instantaneous center of acceleration. The instantaneous center of zero acceleration plays an important role in solving mechanical problems. References Ch H Ch. (1999). nalyses on Instantaneous cceleration Center of Rigi Boies Planner Motion.Journal of Ji mei University(natural Science), Vol. 4,No. 1:19-. Frey R. Zypman. (005). Instantaneous center of rotation an centroes: backgroun an new examples. Int. J. Mech. Enging. Euc., 35 (1): H. E. Williams. (000). note on the use of the instant center as a reference point for angular momentum theorems. Int. J. Mech. Enging. Euc., 8 (): J. F. LI, X. Zhang. (001). Classical Mechanics. Tsinghua University Press, Beijing, 30. K. F. Riley, M. P. Hobson an S. J. Bence. (00). Mathematical Methos for Physics an Engineering (secon eition). Cambrige University Press, Cambrige, 30. SH M CH. (001). Theoretical Mechanics. Higher Eucation Press, Beijing,

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