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Physics 141Mechanics Lecture 11 CM of Rigid BodyYongli Gao To find the center of mass of an object of finite size, we first concentrate on one whose shape.

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Presentation on theme: "Physics 141Mechanics Lecture 11 CM of Rigid BodyYongli Gao To find the center of mass of an object of finite size, we first concentrate on one whose shape."— Presentation transcript:

1 Physics 141Mechanics Lecture 11 CM of Rigid BodyYongli Gao To find the center of mass of an object of finite size, we first concentrate on one whose shape and size are fixed, a rigid body, which is an idealized model of a solid object. We further assume that the object is uniform and of density  =M/V. For a small volume dV at r, the mass is dm=  dV, and the CM of a rigid body is We usually take advantage of the symmetry of the object to reduce the actual calculation. If the object is not uniform, we’d have to find the position dependent density  (r), and keep it as part of the integrand.

2 Example: CM of a Half Sphere Find the CM of a uniform half sphere of radius R 0. Solution: Assume that the z-axis is along the center of the half sphere and perpendicular to the flat bottom, the origion is at the center of the bottom. From the symmetry of the sphere, we have x CM =y CM =0.

3 CM of a Composite Object If an object is composed of a number of pieces, and the CM of each is known, then the CM of the composite object can be obtained by treating each piece as a particle.

4 Example: CM of a Right Angle Ruler Find the CM of a right angle ruler of uniform arms of length l 1 and l 2, respectively. Solution: For each arm, the CM is at its center. x y l2l2 l1l1 +

5 Example: CM of a New Moon Sometime we can use the composite rule to find the CM of each piece. Find the CM of disk of radius R with a hole of radius r cut from its one edge. Solution: The piece cut from the hole is a disk of radius r. The new moon and the disk form a complete disk of radius R, whose CM is at the center. x y

6 Motion of CM The motion of the CM of a particle system or an object of finite size can be described by Newton’s 2nd law as if all the mass is centered at the CM, where is the vector sum of all the external forces acting on the system. The forces exerted by one part of the system on another is called internal forces, a CM is the acceleration of the CM.

7 F i is the total force acting on the ith part of the system, composed of external force F i,ext that is exerted by sources external to the system, and internal force F i,int, that is by the other parts of the system. From the Newton’s 3rd law, The motion of each part of the system, however, is determined by all the forces on it, including both F i,ext and F i,int.

8 Example: Walking on Fishing Boat If you walk slowly from one end of a fishing boat of mass M and length L, how much will the boat move? Suppose your mass is m. Solution: Walking slowly means we can ignore friction by water. Now there’s no net force on the system formed by you and the boat, and the CM remains stationary. The center of the boat is now at m M L x mM x + +

9 Example: Motion of Biatomic Molecule A diatomic molecule can be approximated as two particles of masses m 1 and m 2, linked by a rigid but massless rod of length l. At a given moment, m 1 is moving up with velocity v 10, and m 2 horizontally with v 20, what will be the motion of the molecule as a whole. Solution: m1m1 m2m2 v 10 v 20 l y x

10 The motion of the CM is a projectile of initial velocity of v CM0. The total external force is (m 1 +m 2 )g. The motion of m 1 is determined by the vector sum of m 1 g+F 21, and that of m 2 by m 2 g+F 12 =m 2 g-F 21


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