Physics 2210 Fall smartphysics 09 Work and Potential Energy, Part II 10 Center-of-Mass 10/05/2015

Transcription

1 Physics 2210 Fall 2015 smartphysics 09 Work and Potential Energy, Part II 10 Center-of-Mass 10/05/2015

2 Exam 2: smartphysics units 4-9 Midterm Exam 2: Day: Fri Oct. 09, 2015 Time: regular class time Section 01 12:55 1:45 pm Section 10 02:00 02:50 pm Location: FMAB 13 practice problems for Exam 2 posted on CANVAS and Class Web Page ~woolf/2210_jui/rev2.pdf

3 Unit 09 In your instructor terms: E = j work done by non conservative force j

4 Unit 09

5 Poll A block of mass m, initially held at rest on a frictionless ramp a vertical distance H above the floor, slides down the ramp and onto a floor where friction causes it to stop a distance D from the bottom of the ramp. The coefficient of kinetic friction between the box and the floor is μ k. What is the macroscopic work done on the block by friction during this process? A. mgh B. mgh C. μ k mgd D. 0

6 Example (1/3) The 2 kg block slides down a frictionless curved ramp, starting from rest at a height of h = 4 m. The block then slides d = 10 m on a rough horizontal surface before coming to rest. (a)what is the speed of the block at the bottom of the ramp? (b)what is the energy dissipated by friction? (c) What is the coefficient of friction between the block and the horizontal surface? (%i1) Energy: 0.5*m*v^2 + m*g*y; 2 (%o1) g m y m v (%i2) /* at top from rest */ E0: Energy, v=0, y=h; (%o2) g h m (%i3) /* at bottom of ramp v=v1, y=0 */ E1: Energy, v=v1, y=0; 2 (%o3) 0.5 m v1 (%i4) /* Energy is conserved down the ramp: no friction Solve for v1 */ soln1: solve(e1=e0, v1); rat: replaced 0.5 by 1/2 = 0.5 (%o4) [v1 = - sqrt(2) sqrt(g h), v1 = sqrt(2) sqrt(g h)]

7 Example (2/3) The 2 kg block slides down a frictionless curved ramp, starting from rest at a height of h = 4 m. The block then slides d = 10 m on a rough horizontal surface before coming to rest. (a)what is the speed of the block at the bottom of the ramp? (b)what is the energy dissipated by friction? (c) What is the coefficient of friction between the block and the horizontal surface? (%i5) /* take positive root */ v1: rhs(soln1[2]); (%o5) sqrt(2) sqrt(g h) (%i7) v1, g=9.81, h=4, numer; (%o7) Answer (a) 8.86 m/s (%i8) /* Energy dissipated by friction is the total energy of the system before the rough surface */ E0, m=2, g=9.81, h=4; (%o8) Answer (b) 78.5 J (%i9) /* Work done by surface is W=-E0, but it is also W=-Fs*d where Fs is the magnitude of the friction force which is given by Fs=mu_k*N, but here the normal force has magnitude N=m*g */ N: m*g; (%o9) g m... continued

8 Example (3/3) The 2 kg block slides down a frictionless curved ramp, starting from rest at a height of h = 4 m. The block then slides d = 10 m on a rough horizontal surface before coming to rest. (a)what is the speed of the block at the bottom of the ramp? (b)what is the energy dissipated by friction? (c) What is the coefficient of friction between the block and the horizontal surface? (%i10) Fs: mu_k*n; (%o10) (%i11) soln2: solve(-fs*d=-e0, mu_k); g m mu_k (%o11) [mu_k = -] (%i12) mu_k: rhs(soln2[1]); (%o12) - (%i13) mu_k, h=4, d=10; (%o13) - Answer (c) mu_k = 0.40 h d 2 5 h d

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10 Poll The potential energy of an object U as a function of x looks like the plot shown above. Where is the force the biggest in the negative x direction? a b c d

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12 Unit 10

13 Center of Mass of System of Particles From last class, we defined center-of-mass as a position for a set of point particles X CC m 1x 1 + m 2 x 2 + m 3 x 3 + m 1 + m 2 + m 3 + Y CC m 1y 1 + m 2 y 2 + m 3 y 3 + m 1 + m 2 + m 3 + = 1 N M m ix i i=1 = 1 N M m iy i i=1 Z CC m 1z 1 + m 2 z 2 + m 3 z 3 + m 1 + m 2 + m 3 + = 1 N M m iz i i=1 Where M m 1 + m 2 + m 3 + = i=1 m i is the total mass N

14 Example 10.1 Three particles are located in the xy plane as shown. Find the x- and y= coordinates of the center-of-mass of this system. We have x 1, y 1 = 0.0 m, 0.0 m x 2, y 2 = 1.2 m, 0.0 m x 3, y 3 = 0.0 m, 1.1 m y 1.1 m O m 3 = 2.3 kg m 1 = 1.6 kg 1.2 m m 2 = 1.8 kg x M = m 1 + m 2 + m 3 = 1.6 kg kg kg = 5.7 kg X CC m 1x 1 + m 2 x 2 + m 3 x 3 M = 0.38 m Y CC m 1x 1 + m 2 x 2 + m 3 x 3 M = 0.44m = = 1.6 kg kg 1.2 m kg kg 1.6 kg kg kg 1.1 m 5.7 kg

15 Center-of-mass of uniform and If an object has symmetric objects A. uniform Volume Density of mass (or mass volume density) ρ ( ρ is pronounced rho and is the lower-case Greek letter for r It is often just referred to as density and has units of kg/m 3 ). B. Symmetric shape whose the center is easily located. then The Center-of-mass is located at the center of the object Example: Uniform rod of length L along the x-axis lying between x 1 and x 2 has its center of mass at X CC = 1 2 x 1+x 2 You can calculate the center-of-mass coordinates for a system of finite sized objects by treating them as if they were particles located at THEIR respective centers-of-mass.

16 Poll A glazed doughnut is shown above. Where is the center of mass of this doughnut? A. The center of mass is not defined in cases where mass is missing. B. Somewhere inside the solid part of the doughnut C. In the center of the hole.

17 an old club-ax consists of a symmetrical 7.9 kg stone attached to the end of a uniform 3.2 kg stick. The stick is 98 cm long, and the 18 cm long stone is drilled through its center and mounted on the stick. How far is the center of mass of the club-ax from the handle end of the club-ax? Give your answer in cm. Example 10.2 (1/2) (%i1) /* stick: m1=3.2 kg, of length L1= m=0.98m, stone: m2= 7.9kg length L2=0.18 m */ m1: 3.2; (%o1) 3.2 (%i2) L1: ; (%o2) 0.98 (%i3) m2: 7.9; (%o3) 7.9 (%i4) L2: 0.18; (%o4) 0.18 (%i5) /* CM of stick is at L1/2 */ x1: L1/2; (%o5) continued

18 an old club-ax consists of a symmetrical 7.9 kg stone attached to the end of a uniform 3.2 kg stick. The stick is 98 cm long, and the 18 cm long stone is drilled through its center and mounted on the stick. How far is the center of mass of the club-ax from the handle end of the club-ax? Give your answer in cm. Example 10.2 (2/2) (%i6) /* CM of stone is at L1-L2/2 */ x2: L1 - L2/2; (%o6) 0.89 (%i7) /* total mass */ M: m1 + m2; (%o7) 11.1 (%i8) xcm: (m1*x1 + m2*x2)/m; (%o8) (%i9) /* convert to cm by multiplying 1=100cm/1m */ xcm*100; (%o9) Answer: 77.5 cm

19 Center-of-mass of uniform shapes X CC 1 M lim Y CC 1 M lim N N i=1 N N i=1 x i m i y i m i 1 M 1 M xxx ydd And since we are dealing with a uniform shape then m i = ρ V i Where V i is the volume of the i th piece, and m i dd = ρdd And we have M = ρ dv X CC ρ M x dd, X CC ρ x dd M

20 Center-of-mass of uniform shapes If an object has uniform Volume Density of mass (or mass volume density) ρ, but whose center is not readily identifiable, we resort of integration: Break up a shape into N small pieces whose centers-of-mass ARE identifiable (by symmetry): M m 1, m 2, m 2, m i, m N Each with its center of mass at x 1, y 1, x 2, y 2,, x i, y i,, x N, y N Then we have an approximate formulation for X CC, Y CC : X CC 1 M N i=1 x i m i, Y CC 1 M y m i Which become exact when taking the limit N : X CC 1 M lim N N i=1 x i m i N i=1 xxx