复旦大学：《大学物理》课程教学资源（PPT课件，英文）Chapter 7 Systems of particles

Chapter 7 Systems of particles 7-1 The motion of a complex object 7-2 TWo-particle system 7-3 Many-particle system 7-4 Center of mass of solid objects 7-5 Conservation of momentum in a system of particles

Chapter 7 Systems of particles 7-5 Conservation of momentum in a system of particles 7-4 Center of mass of solid objects 7-3 Many-particle system 7-2 Two-particle system 7-1 The motion of a complex object 

When can the object studied be regarded as a mass point? Doing only translational motion Translational motion Rotational motion Translational Rotational motion Ih Em

When can the object studied be regarded as a mass point? Doing only translational motion Translational + Rotational motion Translational motion Rotational motion

7-1 The motion of a complex object When we projectile a rigid body, the motion of the body looks very complicated We can consider the motion of the rigid body to be a combination of a parabolic trajectory of a"center of mass"(rotational motion is not considered) plus a rotation about center of mass"(translational motion is not considered) How to find the center of mass (CM) of a particle system?

7-1 The motion of a complex object When we projectile a rigid body, the motion of the body looks very complicated. We can consider the motion of the rigid body to be a combination of a parabolic trajectory of a “center of mass” (rotational motion is not considered) plus a rotation about “center of mass” (translational motion is not considered). How to find the center of mass (CM) of a particle system?

7-2 TWO-particle system Do an experiment to find cm in a two-particle system As an example we consider a baton consists of two particles m, m, at its ends a and B connected by a thin rigid rod of fixed length and negligible mass B

7-2 Two-particle system As an example, we consider a baton consists of two particles , at its ends A and B, connected by a thin rigid rod of fixed length and negligible mass. m2 = 2m1 m1 m2 m1 m2 A B Do an experiment to find CM in a two-particle system

We give the rod a push along the frictionless horizontal surface and examine its motion Snapshots of the locations of points a and B at successive intervals of time Clearly both m, and m2 are accelerated, however, one point in the rod(point c moves with constant velocity

Clearly both and are accelerated, however, one point in the rod (point C) moves with constant velocity. m1 m2 We give the rod a push along the frictionless horizontal surface and examine its motion. Snapshots of the locations of points A and B at successive intervals of time

If point c is regarded as a reference, A and B points rotate with a constant rotational speed 3 View the motion from the reference of point C So point c is actually the center of mass

If point c is regarded as a reference, A and B points rotate with a constant rotational speed. View the motion from the reference of point C. So point c is actually the center of mass

Fig 7-5 By building a Cartesian coordinate, position of y point c is found at m11+m C C cn (7-1) m1+m2 or written as mx,+.x cm m1+m2 A (7-2) m,y1+m2y B m1+m2

Fig 7-5 m1 m2 1 r  x y O 1 2 1 1 2 2 m m m r m r rcm + + =    1 2 1 1 2 2 m m m x m x xcm + + = 1 2 1 1 2 2 m m m y m y ycm + + = cm r  2 r  ' cm r  C C’ By building a Cartesian coordinate, position of point c is found at : (7-2) (7-1) or written as:  m1 m2 A B C

From Eq(7-1), the velocity and acceleration of the cm are: m, VIt m2 1 cn 21+m (7-4) +m a (7-6) m,+m Can we also find cm is zero from Eg(7-6? 11a1+m2a E+ F 0

From Eq. (7-1), the velocity and acceleration of the CM are: (7-4) (7-6) 1 2 1 1 2 2 m m m v m v dt d r v cm cm + + = = → → → → 1 2 1 1 2 2 m m m a m a dt d v a cm cm + + = = → → → → Can we also find is zero from Eq. (7 cm -6)? a  → → → → + = F r + F r m a m a 1 1 2 2 1 2 = 0 → c a → → F2r = − F1r

How about the motion if the system has net external forces? Suppose there is an external force on each particle in above expt, then n,,+m a 2=∑F+∑ f=F+ftf+e 2 2r F+F2=0, and if write∑Fa+∑F2a=∑F ∑Fn n1a+mmoa ntn Newton's second law for ∑ Fen=(m,+m2)acm systems of particles This looks very like a particle of mass m, +m, located the center of mass

How about the motion if the system has net external forces? Suppose there is an external force on each particle in above expt., then This looks very like a particle of mass located at the center of mass. → → → → → → → → m1 a1 + m2 a2 = F1 +F2 = F1ext + F1r + F2ext + F2r 0 1 + 2 = → →  F r F r    → → → F1ext + F2ext = Fext , and if write → →  ext = + cm F (m m )a 1 2 m a + m a =Fext    1 1 2 2         + + = 1 2 1 1 2 2 m m m a m a acm    m1 m2 + Newton’s second law for systems of particles

7-3 Many-particle system Consider a system consisting of N particles of masses m1m2……mx: The tota mass Is M=∑ (7-10) Each particle can be represented by its location rn, velocity v, and its acceleration a The CM of the system can be defined by logical extension of Eq(7-1:

7-3 Many-particle system Consider a system consisting of N particles of masses …… . The total mass is (7-10) Each particle can be represented by its location , velocity and its acceleration . The CM of the system can be defined by logical extension of Eq(7-1): m1 m2 mN M =mn → n v n r  n a 