{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Dynamics Homework 2 - ENGINEERING MECHANICS Dynamics W.F...

Info iconThis preview shows pages 1–7. Sign up to view the full content.

View Full Document Right Arrow Icon
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Background image of page 2
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Background image of page 4
Background image of page 5

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Background image of page 6
Background image of page 7
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: ENGINEERING MECHANICS - Dynamics W.F. Riley & L.D. Sturges 13-92 A shore cannon is firing at a pirate ship 560 m offshore. If the hutial velocity of the cannon ball is v0 = 80 m/s and its acceleration . 2 . is 9.81 m/s , downward, determine a. The initial angle 90 at which the cannon must be fired. The time T that the pirate ship will have to evade the cannon ball. Solution The acceleration of the cannonball is _ = -9.81 3 m/s2 Integrating the acceleration gives the velocity v = so cos 90 ’1‘ + (80 sin 90 - 9.8lt) 3 m/s where the constant of integration has been chosen to match the initial velocity of 80 m/s 1 90. Integrating one more time gives the position —‘ A 2 A r 80t cos 90 1 + (80t sin 90 — 4.905t ) J m The cannonball reaches the pirate ship when x = 80t cos 60 = 560 m 2 y 80t sin 90 - 4.905t = O m Solving Eq. a for t 7/cos 90 and substituting into Eq. b gives 80 sin 90 4.905(7/cos 90) sin 290 = 0.85838 0 O - 29.567 ENGINEERING MECHANICS - Dynamics W.F. Riley & L.D. Sturges 13-94* A daredevil jumps his motorcycle across a canyon as shown. If 2 the acceleration of the daredevil is 9.81 m/s , downward, determine The initial speed vO required to complete the jump. The landing speed vf. h0 = 70.0 m The surface angle 9 required at the end of the jump to ensure a smooth landing. Solution The acceleration of the cyclist -" A 2 a = -9.81 J m/s Integrating the acceleration gives the velocity .. V = V0 3 - 9.81t 3 m/s where the constant of integration has been chosen to match the initial velocity of V0 -9. Integrating one more time gives the position _. A 2 A r = Vot 1 + (35 - 4.905: ) J m 2 The cyclist completes the jump when y = 35 - 4.905t t = 2.671 s at which time x = Vot = 100 m Therefore, v0 = 37.44 m/s 3 134.8 km/h Also, at t 2.671 s the velocity of the cyclist is v v = 37.44 m/s x 0 vy -9.81t = -26.20 m/s v 45.7 m/s E 34.99° 164.5 km/hr E 34.990 ENGINEERING MECHANICS - Dynamics W.F. Riley & L.D. Sturges 13-100 In Example Problem 13—9 it was shown that the initial angle 60 much gives the maximum range of a projectile is 45°. Although this result holds for any given initial speed, it does require that the initial and final heights are the same. Determine the initial angle 90 which gives the maximum range of a projectile if the final height is above the initial height as shown. Solution The velocity and position of the projectile are obtained by integrating the acceleration A . 2 A = Vot cos 90 1 + (Vot sin 90 gt /2) J where the constants of integration have been chosen to satisfy the initial velocity (V = v0 cos 90 + V0 sin 90 j) and initial position (F = 6) when t = O s. The projectile hits the upper surface when 6 = x Vot cos 0 R . 2 _ y vot Sin 90 gt /2 — h Solving Eq. b for t . 2 _ 2 2v0 Sin 60 + I 4V0 Sin 90 89h 29 and substituting into Eq. a gives an expression for the range R as a function of initial angle 60 . 2 . 2 2v0 Sin 90 + V 4v0 Sin 90 89b 29 cos 60 (Problem 13-100 continues ...) ENGINEERING MECHANICS - Dynamics W.F. Riley & L.D. Sturges (Problem 13-100 - cont.) which can be rewritten 2 R - v2 sin 29 + v cos 6 V 4 2 s' 29 — 8 b g— o o o 0 v0 m o 9 For maximum range, the derivative dR/d90 = O 2 2 2 6 — ' 9 I ‘ — 2V0 cos 2 0 v0 Sin 0 4V0 Sin 90 89b 2 4v s'n 6 c 6 o l 0 OS 0 2 , 2 V 4V0 Bin 60 89h which after a little algebra and trigonometry 2 2.2 3,3 2V0 cos 290 V 4VO Sin 90 89h 4V0 Sin 9 3 2 + ' + 4 ' — av 91') 5111 6 V 511') 6 COS 6 0 /2,2 _2_ , vO cos 260 4v0 Sin 90 8gb — 2vo Sin 90 cos 290 4gb Sln 90 2_2 2 ,2 2 2 _ 16(gh) Sin 60 + 16vogh Sin 90 cos 290 + 8vogh cos 260 — O 29 dR/de0 [I O + v cos 9 O 0 0 ON ,2 _2 _ 29h Sin 90 + v cos 290[2 Sin 90 + cos 290] — 0 cos 260 = 0 ON 2 29h sin 90 + .2 2 .2 _ 2gb Sin 90 + v0 (cos 90 Sin 90) — O 2 v0 - 29h 2 . . . . Note that v0 must be greater than 29h. That is, if the progectile were 0 fired vertically upward (90 = 90 ) with an initial velocity of V0, it 2 0 would rise to a maximum height of vo/Zg. If 90 < 90 , then the maximum 2 height will be less than Vo/2g. But the maximum height must be greater 2 than the height of the upper surface; vo/Zg > h. ENGINEERING MECHANICS — Dynamics W.F. Riley & L.D. Sturges 13-107* A radar tracking an airplane gives the coordinates of the plane as r(t) and 9(t). During some segment of the tracking, the radar recorded the following values: t rstz 9st! 8,020 ft 29.9: 0 2 8,590 ft 28.70 4 9,285 ft 27.60 6 10,110 ft 26.5 For this segment of the flight, a. Estimate the velocity components vr and V6' b. Estimate the acceleration components at and a6. Solution Since the position data is given at discrete values of time, the derivatives must be approximated. For example, - A - o r(1) z ‘Kf' = —§§29—3—§93—— = 285 ft/s Similarly, -- A. . — 2 r(2) e A: = —331—§§——3§§— = 31.25 ft/s Therefore, completing the table t r r r 9 6 6 0 0.52185 1 -0.01047 2 0.50091 0.00044 3 -0.00960 4 0.48171 0.00000 5 -0.00960 6 0.46251 gives the estimates at t = 3 s: r E 8938 ft, r E 347.5 ft/s, r E 31.75 0 2 o ft/s , 6 2 0.49131 rad = 28.15 , 6 2 —0.00960 rad/s, and 6 z 0.00022 2 . . rad/s . Then, the veloc1ty and acceleration components are Vr = roz 347 ft/s .............................. Ans. v6 = re a :85.8 ft/s .............................. Ans. ar = ;"_ r92.3 30.9 ft/s2 .............................. Ans. a9 = r9 + 2£6 a -4.71 ft/s2 .............................. Ans. V a 358 ft/s a 14.280, 5 a 31.3 ft/s2 a 19.490 ENGINEERING MECHANICS — Dynamics W.F. Riley & L.D. Sturges 13-119* A car drives over the top of a hill that has a radius of curvature of 110 ft. If the normal component of acceleration necessary to keep the car on the road becomes greater than that provided by gravity, the car will become airborne. Determine the maximum constant speed V at which the car can go over the hill. Solution 5 9 = / (110)(32.2) ft/s = 40.6 mi/h ENGINEERING MECHANICS — Dynamics W.F. Riley & L.D. Sturges 13-122 The roller coaster cars shown are traveling at a speed of v = 50 km/h (and {I = 0) when they pass over the top of the hill. If the radius of curvature is p = 21 m, determine the acceleration of the cars as they pass over the top of the hill. Solution 50 km/h = 13.889 m/s 2 . 2 a =_=_13_889_=9.19m/s n p 21 Therefore the total acceleration is 2 2 2 = + = . Jr a at an 9 19 m/s ...
View Full Document

{[ snackBarMessage ]}