Linear Momentum Center of Mass Lana Sheridan De Anza College Nov 14, 2017
Last time the ballistic pendulum 2D collisions center of mass finding the center of mass
Overview center of mass examples center of mass for continuous mass distributions
Center of Mass Last lecture we introduced the center of mass, the average point of the mass for a collection of particles: r CM = 1 m i r i M i This is a vector equation, so we can also express it as a set of component equations, one for each dimension.
Example 9.10 - Three technique Particles for determining athe plane es Figure 9.16 An experimental center of gravity of a wrench. A system consists of three particles located as shown. Find the center of mass of the system. The masses of the particles are m 1 = m 2 = 1.0 kg and m 3 = 2.0 kg. 8. Find the cenm 2 5 1.0 kg and y (m) 3 2 m 3 Example 9.10) Two ated on the x axis, ticle is located on wn. The vector indin of the system s 1 0 S r CM m 1 m 2 1 2 x (m) 3 he center of mass developed in this section. 1 Serway & Jewett, page 269.
Example 9.10 2 dimensions: can find the x and y coordinates of the center of mass separately. Total mass, M = 2(1.0 kg) + 2.0 kg = 4 kg. x-direction: x CM = 1 m i x i M i = 1 M (m 1x 1 + m 2 x 2 + m 3 x 3 ) = 1 4 (1 1 + 1 2 + 2 0) = 0.75 m
Example 9.10 x-direction: x CM = 0.75 m y-direction: y CM = 1 m i y i M i = 1 M (m 1y 1 + m 2 y 2 + m 3 y 3 ) = 1 (1 0 + 1 0 + 2 2) 4 = 1.0 m
Example 9.10 x-direction: x CM = 0.75 m y-direction: y CM = 1 m i y i M i = 1 M (m 1y 1 + m 2 y 2 + m 3 y 3 ) = 1 (1 0 + 1 0 + 2 2) 4 = 1.0 m Putting both components together: r CM = (0.75 i + 1.0 j) m
Continuous mass distribution On a microscopic scale we don t really believe mass is continuously distributed. (Atoms, etc.!) However, on a macroscopic scale it sure seems like it is, and its a very good approximation. It is also much easier than trying to describe every last atom / subatomic particle in a solid. We need to revise our previous center of mass (CM) definition to suit this case: 1 r CM = lim r i m i m i 0 M r CM = 1 r dm M i
Continuous mass distribution r CM = 1 r dm 9.6 The Center of Mass M 269 xtended object in (9.34) An extended object can be considered to be a distribution of small elements of mass m i. y.32 and 9.33. lies on an axis of r of mass of a uniass of a sphere or s, each small mass all these forces is z S r i S r CM m i CM x
Discrete vs Continuous mass distribution Compare a discrete and continuous mass distribution along 1 dimension. Assume all particles have equal mass and each are separated by one meter. Where is the center of mass of this distribution? y x
Discrete vs Continuous mass distribution Compare a discrete and continuous mass distribution along 1 dimension. Assume all particles have equal mass and each are separated by one meter. Where is the center of mass of this distribution? y x
Discrete vs Continuous mass distribution y x The calculation would be: x CM = 1 m i x i M i
Discrete vs Continuous mass distribution y x The calculation would be: x CM = 1 m i x i M = 1 6m = 2.5 m i 5 m i The center of mass is at the midpoint of the line of point masses. i=0
Discrete vs Continuous mass distribution.75 i^ 1 1.0 j^2 m Center of mass of a rod of uniform density. Density, ρ = M V, is mass per unit volume. Here however, we will consider a rod that can be treated as 1 dimensional. We need the mass per unit length: λ = M L. lies midway ength. y L dm = l dx.19, so y CM 5 ause we need Figure 9.19 (Example 9.11) The tion in Where Equa- is the center geometry of mass? used to find the center of mass of a uniform rod. x dx x
Discrete vs Continuous mass distribution.75 i^ 1 1.0 j^2 m Center of mass of a rod of uniform density. Density, ρ = M V, is mass per unit volume. Here however, we will consider a rod that can be treated as 1 dimensional. We need the mass per unit length: λ = M L. lies midway ength. y L dm = l dx.19, so y CM 5 ause we need Figure 9.19 (Example 9.11) The tion in Where Equa- is the center geometry of mass? used to find the center of mass of a uniform rod. x dx x
Discrete vs Continuous mass distribution How to do the calculation: x CM = 1 x dm M = 1 x dm M Observe that dm = λ dx: x CM = 1 x(λ dx) M = 1 L M λ x dx 0
Discrete vs Continuous mass distribution Observe that dm = λ dx: x CM = 1 x(λ dx) M = 1 L M λ x dx 0
Discrete vs Continuous mass distribution Observe that dm = λ dx: x CM = 1 x(λ dx) M = 1 L M λ x dx 0 = 1 ( ) L 2 M λ 2 0 = 1 ( ) M L 2 M L 2 = L 2 In this case (uniform density) the center of mass is at the center of the rod.
established with a plumb bob) is drawn when the wrench has stopped swinging. The wrench is then hung from point C, and a second vertical line CD is drawn. The center of gravity is halfway through the thickness of the wrench, under the intersection of these two lines. In general, if the wrench is hung freely from any point, the vertical line through this point must pass through the center of gravity. Center of Mass Intuition Question Quick Quiz 9.7 1 A baseball bat of uniform density is cut at the location of its center of mass as shown. Which piece has the smaller mass? (d) impossible to determine Q uick Quiz 9.7 A baseball bat of uniform density is cut at the location of its center of mass as shown in Figure 9.17. Which piece has the smaller mass? (a) the piece on the right (b) the piece on the left (c) both pieces have the same mass Figure 9.17 (Quick Quiz 9.7) A baseball bat cut at the location of its center of mass. (A) the piece on the right Example 9.10 (B) the piece on the left The Center of Mass of Three Particles A system consists of three particles located as shown in Figure 9.18. Find the center of mass of the system. The masses of the particles are m 1 5 m 2 5 1.0 kg and m 3 5 2.0 kg. (C) both pieces have the same mass (D) impossible to determine SOLUTIO N y ( 3 2 m Conceptualize Figure 9.18 shows the three masses. Your intuition should tell you that the 1 Serway & Jewett, page 269. Figure 9.18 (Example 9.10) Two particles are located on the x axis, and a single particle is located on 1
established with a plumb bob) is drawn when the wrench has stopped swinging. The wrench is then hung from point C, and a second vertical line CD is drawn. The center of gravity is halfway through the thickness of the wrench, under the intersection of these two lines. In general, if the wrench is hung freely from any point, the vertical line through this point must pass through the center of gravity. Center of Mass Intuition Question Quick Quiz 9.7 1 A baseball bat of uniform density is cut at the location of its center of mass as shown. Which piece has the smaller mass? (d) impossible to determine Q uick Quiz 9.7 A baseball bat of uniform density is cut at the location of its center of mass as shown in Figure 9.17. Which piece has the smaller mass? (a) the piece on the right (b) the piece on the left (c) both pieces have the same mass Figure 9.17 (Quick Quiz 9.7) A baseball bat cut at the location of its center of mass. (A) the piece on the right Example 9.10 (B) the piece on the left The Center of Mass of Three Particles A system consists of three particles located as shown in Figure 9.18. Find the center of mass of the system. The masses of the particles are m 1 5 m 2 5 1.0 kg and m 3 5 2.0 kg. SOLUTIO N (C) both pieces have the same mass (D) impossible to determine y ( 3 2 m Conceptualize Figure 9.18 shows the three masses. Your intuition should tell you that the 1 Serway & Jewett, page 269. Figure 9.18 (Example 9.10) Two particles are located on the x axis, and a single particle is located on 1
Continuous mass distribution: Another Example Center of mass of a cylinder of uniform density, ρ. z dr r R L 10.16 (Exam- ) Calculating I e z axis for a solid cylinder.
Continuous mass distribution: Another Example Let s choose our axes so that z points along the length of the cylinder, and the origin is right in the center of the cylinder. Probably, you can easily guess where the CM will be.
Continuous mass distribution: Another Example Let s choose our axes so that z points along the length of the cylinder, and the origin is right in the center of the cylinder. Probably, you can easily guess where the CM will be. At the origin!
Continuous mass distribution: Another Example How could we prove it?
Continuous mass distribution: Another Example How could we prove it? Along the z-direction this cylinder is very similar to the thin rod we just considered. Observe that dm = (πr 2 )ρ dz: z CM = 1 z(πr 2 ρ dz) M = (πr2 )ρ πr 2 Lρ L/2 L/2 z dz
Continuous mass distribution: Another Example How could we prove it? Along the z-direction this cylinder is very similar to the thin rod we just considered. Observe that dm = (πr 2 )ρ dz: z CM = 1 z(πr 2 ρ dz) M = (πr2 )ρ πr 2 Lρ = 0 L/2 L/2 z dz
Continuous mass distribution: Another Example Along the x-direction this cylinder we are looking down on a circle. Let x = r cos θ, y = r sin θ Observe that dm = (2yL)ρ dx: (Looking at just the positive x part of the cylinder, viewed top down.) 1 Or, leave the integral in terms of x and use the substitution u = x 2.
Continuous mass distribution: Another Example Along the x-direction this cylinder we are looking down on a circle. Let x = r cos θ, y = r sin θ Observe that dm = (2yL)ρ dx: x CM = 1 x(2yl)ρ dx M = = = 0 Lρ πr 2 L ρ πr 2 0 π r 2 (2 sin θ cos θ)( r sin θ dθ) [ 2 3 r 3 sin 3 θ ] 0 π 1 Or, leave the integral in terms of x and use the substitution u = x 2.
Continuous mass distribution: Another Example By symmetry the evaluation for y CM must go exactly the same way as for x CM, therefore, y CM = 0. This is what we expected all along, but this same technique can be used in cases where we cannot guess the answer.
Summary center of mass for continuous mass distributions 3rd Collected Homework! due Monday, Nov 20. Next Test Monday, Nov 27. (Uncollected) Homework Serway & Jewett, Ch 9, onward from page 288. Probs: 45, 47, 48, 49, 50 Look at example 9.12 and make sure you understand it. Read Chapter 9, if you haven t already.