The Classical Mechanics - Speech or Presentation Example

Chapter 2 No. 10 Solution: this is a projectile motion question requiring the illustration by a figure as shown. The figure illustrates initial conditions & acceleration in the problem. Since this involves a 2 dimensional problem, the solution would involve assigning initial position, initial velocity, and acceleration are specified by two numbers: xo = 0 vox = vo cos 60 = (.5) vo ax = 0 . yo = 0 voy = vo sin 60 = (.866) vo ay = ‑ 32 ft/s2 . x(t) = (.5) vo t y(t) = ‑ (1/2)(32) t2 + (.866)vo t vy(t) = ‑ 32 t + (.866)vo . Making use of other information given to get to the solution considering that Vo is not given. let the time at which the projectile strikes the building be t, as shown in the figure. Since vo is not given in the problem, some other piece of information must be given. This gives: 03-3 x(t) = 80 = (.5) vo t y(t) = 48 = ‑(1/2)(32)t2 + (.866)vo t . Hence we have 2 equations in 2 unknowns and can solve for both t and vo. Solving we find: t = 2.38 seconds; vo = 67.3 ft/sec. vx = vox = (.5)(67.3) = 33.6 ft/sec vy(t=2.38s) = ‑ 32(2.38) + (.866)(67.3) = ‑ 17.9 ft/sec . The following figure illustrates the velocity vector of the component projectile motion. The figure helps in determining the magnitude and direction of the particle.: v = = 38.1 ft/sec. tan  = (17.9)/(33.6)  = 28o (below hor. as shown) Chapter 3 No. 24 x-direction y-direction Neglecting the height , assume x=y=0 at t=0,then , & The speed and total displacement as functions of time are found to be By determining the value of x when the projectile falls back to ground that is when y=0, helps in finding the range. one value of y=0 occurs for t=o and the other one for t=T the Range R is found from Adding the effect of air resistance to the projectile motion will lead to the reduction in the range, as included in the assumption considering that the resistance is directly proportional to the motion of the projectile. The initial conditions are the same as above initial case Then the equations of motion, become The solution is and Chapter 4 No. 10 Solution: solving the problem would involve designing a setup as shown in the figure below. The initial conditions are determined followed by determining the acceleration after the CS is chosen. The initial conditions: xo = 0 ; vox = vo cos  ; yo = 7 ft ; voy = vo sin  . The acceleration is given by: ax = 0 ay = ‑ 32 ft/s2 . Inserting these values into the general equations of motion in 2‑dimensions, will lead to: 03-4 x(t) = vo cos  t ; y(t) = ‑ (1/2)(32) t2 + (vo sin  ) t + 7 ; vy(t) = ‑ 32 t + vo sin  . At t = 1.5 seconds we have: x(1.5s) = 30 = vox (1.5) y(1.5s) = 10 = ‑ 16(1.5)2 + voy(1.5) + 7 Thus: vox = vo cos  = 20 ft/sec; voy = vo sin  = 26 ft/sec. The magnitude & direction of the initial velocity is then: vo = = 32.8 ft/sec; tan  = (26)/(20)  = 52.4o (That is, 52.4o above the horizontal). The maximum height above the floor occurs at a time t when vy(t) = 0. Hence: vy(t) = 0 = ‑ 32 t + 26  t = .866 sec. Then y(0.866sec) = ‑ 16 (.866)2 + (26)(.866) + 7 = 17.52 ft. Chapter 5 No. 5 Solution: this is a projectile motion problem hence solving the problem would involve designing a setup as shown in the figure below. The initial conditions are determine followed by determining the acceleration after the CS is chosen. The initial conditions: x0 = 0; y0 = 0; v0x = v0 cos 37; v0y = v0 sin 37 The acceleration is: ax = 0; ay = ‑ 32 ft/sec2. 03-2 In 2 dimension, the following is the general equations of motion for constant acceleration: x(t) = (1/2) ax t2 + vox t + xo y(t) = (1/2) ay t2 + vox t + yo vx(t) = ax t + vox vy(t) = ay t + voy Inserting the known values to get the unknown x(t) = (48)(4/5) t y(t) = ‑ (1/2)(32) t2 + (48)(3/5) t vy(t) = ‑ 32 t + (48)(3/5) y(t) = 0 = ‑ 16 t2 + (48)(3/5) t  t = 0, or t = 1.8 sec. Hence the particle will land at x(t) = x(1.8s) = (48)(4/5)(1.8) = 69 ft from the origin. For the second player initial position (t=0) was 100 ft from the origin, and needs to reach a point 69 ft from the origin time to be taken is in 1.8 sec for catching the ball. This gives an average velocity of vave = (x2 ‑ x1)/(t2 ‑ t1) = (69 ‑ 100)/(1.8) = ‑ 17 ft/sec. the negative sign shows the direction that the player needs to run to and in this case it is towards the origin. Chapter 6 No. 12 Solution: the following shows a figure drawn to represent the setup with the CS chosen and initial conditions determined. The following are the initial conditions: xo = 0 ; vox = vo cos  ; yo = 7 ft ; voy = vo sin  . The acceleration in the problem is: ax = 0 ay = ‑ 32 ft/s2 . the values are then inserted in the general equation for 2D 03-4 x(t) = vo cos  t ; y(t) = ‑ (1/2)(32) t2 + (vo sin  ) t + 7 ; vy(t) = ‑ 32 t + vo sin  . At t = 1.5 seconds we have: x(1.5s) = 30 = vox (1.5) y(1.5s) = 10 = ‑ 16(1.5)2 + voy(1.5) + 7 Thus: vox = vo cos  = 20 ft/sec; voy = vo sin  = 26 ft/sec. The magnitude & direction of the initial velocity = vo = = 32.8 ft/sec; tan  = (26)/(20)  = 52.4o this means that it is 52.4o above the horizontal The maximum height above the floor occurs at a time t when vy(t) = 0. Hence: vy(t) = 0 = ‑ 32 t + 26  t = .866 sec. Then y(0.866sec) = ‑ 16 (.866)2 + (26)(.866) + 7 = 17.52 ft. Chapter 8 No. 27 the expanded form for the equation of x ) Since, then the approximate range will be The quantity 2UV/g can be written as Neglecting the air resistance will lead to the recognition of the range R. Therefore , Meaning the expansion will not converge unless which can be written as Here β=b/2m represents the damping parameter and ω0 =represents the characteristic angular frequency when there is no damping. The following equations represent the roots of the auxilliary The general solution of the equation is : ] the differential equation for perturbation method: , for definiteness, the initial conditions are y(0)=0,y’(0)=1. now for the second order the expression reduces to: they break into y j (0) =0 because the initial conditions don’t depend on ε ,so yo’(0)=1, yo’(0)=1 ,y2 ‘(0)=0 The solution for Substituting the above equation to the equation for y1 : homogeneous solution For the second –order term we get the equation : function -……………………… To judge this approximation ,comparing it with the exact solution .It is this is the desired approximate expression for the flight time . Chapter 9 No. 39 Solution: the following figure shows the setup of the motion to help in solving the question. The figure shows a chosen CS, have written down the initial conditions (initial position & velocity) of the shell (at t=0). xo = 0 vox = vo cos 60 = (.5)(2000) = 1000 ft/sec ax = 0 . yo = 0 voy = vo sin 60 = (.866)(2000) = 1800 ft/sec ay = ‑ 32 ft/s2 . The general equations of motion for constant acceleration in 2‑dimensions are: x(t) = (1/2) ax t2 + vox t + xo y(t) = (1/2) ay t2 + vox t + yo vx(t) = ax t + vox vy(t) = ay t + voy the known values for acceleration & initial conditions are inserted followed by obtaining the specific equations for the shell: x(t) = 1000 t ; y(t) = ‑ (1/2)(32) t2 + 1800 t ; vy(t) = ‑ 32 t + 1800 . 03-5 vy(t) = 0 = -32 t + 1800  t = 56.3 sec. Then: ymax = y(t) = -(1/2)(32)(56.3)2 + 1800 (56.3) = 50,625 ft. This means that there is possibility of reaching the elevation of the planbe. Consequently, this helps in determining the time at which the plane will reach an elevation of 42,000 ft)? y(t) = 42,000 = ‑ 16 t2 + 1800 t ‑‑> t2 - 112.5 t+ 2625 = 0 This requires the use of a quadratic equation to reach to the solution. Using the quadratic formula we have: t = = t1 = 33.03 sec; t2 = 79.47 sec. First time: The distance at the first time x(t=33.03) = 33,030 ft (6.256 mi) from the origin. Second time: At the 2nd time x(t= 79.47) = 79,470 ft (15.05 mi). Hence the shell cannot hit the plane, since the plane is travelling at 880 ft/sec thereby covering a distance of 13.245 mi in 79.47 seconds. xB(t) = 880 t yB(t) = -(1/2)(32) t2 + 42,000 . Hence, the time to fall" is: yB(t) = 0 = -16 t2 + 42,000  t = 51.23 sec. At this time, 51.23 sec the bomb covers a horizontal distance of = (880)(51.23)/5280 = 8.54 mi. It will be shot down 11.76 miles away! Read More