Exam1-soln-MW

Exam1-soln-MW - Problem 1. This problem is most...

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Unformatted text preview: Problem 1. This problem is most conveniently solved by using the direct definition of acceleration, viz., a = dv/dt. Once the velocity is known, we solve for position by integrating dx/dt = v . . . ....... ... ... ... ...... ... •..... . ... .... .... ... ..... ... ... ..... ..... ..... ... ..... ..... ..... . ... ... .. ... ... ... .. .. .. ... ... ... ... ... ..... ... ... ... ..... ... ... . . ... ... ... ... ... .. ... .. ... ... ... ... ... ..... ... ..... ... ... . ... ... ... ... ... ... ... ... ... .... . ... ... ... ... ... ... ... .... . ... ... ... ... ... ... ... .... ... ... ... ... . ... ... ... ... ... ... ... .... . ... ... ... ... ... ... ... .... . ... ............................................. ... ............................. .......... ... . ... . ... ... ... ..................................................... ... ................................................ ... .... . . ... ............................................... . . ... ................................. . ... ... ... ... ... ........................................... . ... 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A x Solution: For this problem, we will have to determine the slider’s position as a function of time. Hence, from the definition of acceleration, we have 12 dv vo − v 2 dt =⇒ dv = a dt = − a= dt τ Rearranging terms, we have dv dt dt v =− =⇒ d sin−1 =− 2 τ vo τ vo − v 2 Then, integrating and noting that v (0) = 0, there follows v t t =⇒ v = vo sin − sin−1 =− vo τ τ Thus, the slider’s velocity as a function of time is t v (t) = −vo sin τ We can now determine the slider’s position by using the definition of velocity, viz., dx t =⇒ dx = v dt = −vo sin v= dt dt τ Integrating and using the fact that x(0) = , t t dt = + vo τ cos τ τ 0 Therefore, the slider’s position as a function of time is t x(t) = + vo τ cos −1 τ Now, we know that when t = π τ , the slider’s position is x = 1 . So, 3 2 t t= t t=0 x(t) = − vo sin 1 π 1 = + vo τ cos − 1 = − vo τ 2 3 2 Therefore, solving for vo yields vo = 1 τ =⇒ = vo τ Problem 2. To solve this problem, we first solve for the projectile’s position as observed by a stationary observer. Then, we compute the accelerating vehicle’s position, again as observed by a stationary observer. Equating the projectile and vehicle positions yields the desired equation. g . . . . . . . . . . . . . . . . . . . . . ... ... .. . . .. .. . ......... . . . .. . . . . . . . . . . . .. . . . ... .. ... ..... . . . .. .. ..... . . . . . .. .. .. . .. ... . . ..... . .. .. ..... . .... .. .. .. ... .. .... .. .. .. ... .... .. .. .. ... . ... .. . . . . .. .. ................... ................... . . .. . ............................................... .. . .. . . . . ...... . . .......................................... .................................................. ......... ............................................................... ............ .......... . . .. . ..... . ... . . ... ...... . . .. . . .... ... .. . ....... ............................................................................ .. .................. .............................................. . ............ ............... .. .......... ......................................................... ......... ...... ... ..................... .. .. .. . .. . .. ................................ . .. ........ .. ............................................... .... .......... ..... ............ . ........... . . .. . ............................................. ....... . . ... . ..... ..................................... ........ ......................................... ............. ... . ........... . . ............ ................................................................................................................................................................................................................................................................................... .... .. .. . ................................................................................................................................................................................................................................................................................ ........ ......... nV α • a = 1g 2 • vx (t) ........................................................................................................................................................................................................................................................................................ .... .................. .... ............. .... .......... ....... .... ............. .... ............ ..... .... ............. . ........................................................................................................................................................................................................................................................................................ ........................................................................................................................................................................................................................................................................................ v v Solution: Denote the ball’s coordinates by xB and zB . Then, the equations of motion and initial conditions are d2 xB = 0; xB (0) = 0, xB = nV cos α ˙ dt2 d2 zB = −g ; zB (0) = 0, zB = nV sin α ˙ dt2 The solution is 1 xB (t) = nV t cos α and zB (t) = nV t sin α − gt2 2 The projectile strikes the vehicle when zB = 0, i.e., when 1 nV t sin α = gt2 2 =⇒ t= 2nV sin α g Thus, when the projectile strikes the vehicle, its horizontal coordinate, xB , is xB = 2n2 V 2 sin α cos α g Denoting the vehicle’s position by xA , its motion is governed by 1 d2 xA = g; 2 dt 2 The solution is xA (0) = L, xA (0) = V ˙ 1 xA (t) = L + V t + gt2 4 When the particle strikes, the vehicle’s position is 2nV sin α 1 2nV sin α xA = L + V +g g 4 g Simplifying, we find xA = L + So, xA = xB when 2nV 2 sin α n2 V 2 sin2 α 2n2 V 2 sin α cos α L+ + = g g g 2 2nV 2 sin α n2 V 2 sin2 α + g g 2 which can be rearranged to yield V2 2V 2 sin α sin α(2 cos α − sin α)n2 − n−L=0 g g Multiplying through by g/(2V 2 ), the equation from which n can be determined is sin α(2 cos α − sin α)n2 − (2 sin α)n − gL =0 V2 3 Problem 3. Note that, with θ defined as indicated in the figure, positive velocities and forces are in the clockwise direction. z g z R θ R z ...................................................................................................................................................................... ........................................................................................................................................................................... .. . .. ... .. .. . . ... .. ..... ..... .. ....................................................... ...... ................................................................................ ........................................................ .. . ................................................................................................................................................................................... ................................................................................................................................................................................................ ................................................................................... ...... ...... .................................. .. .. . . . . . . . .................................................................................................................................................................................... . . .. . .. . . . . . . . . . . . . .. ........................................................................................................................................................................................................................................................................... . . .. . . . . . . . . . . . . . . . . . . ... .. ........................................................................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................................................................... .................................................................... . ............ ..... ............................ .. ................................. . . . . . . . . . . ... . . ................................................................................................................................................................................. ........................................................................................................................................ . .................................................................. . .................................................................................. .......................................... ....... ..................... ....... ....... ..................... ..................... .. . . . ............. . .. . ........................................................................................................................................................................................................................................................................................................................................................................................................ .................................................................................................................................................................................................................................................................................................................................................... ................................................................... .... . .... .... .... ..... .... .... .... .... .... .... .... ..... ..... ......... .... ...... . .. .. . . .. .. .. . . .. . . (a) Newton’s Second Law tells us that ˙ Fr = m r − rθ2 ¨ and ¨ F θ = m r θ + 2r θ ˙˙ Up to the point of separation, the ball moves on the cylindrical surface, wherefore r = R and r = r = 0. Therefore, ˙¨ ˙ Fr = −mRθ2 and ¨ Fθ = mRθ There are two forces acting on the sphere, viz., the ball’s weight, mg , and the normal reaction force, N . As shown below, mg acts vertically downward and N acts radially outward. N θ mg z mg sin θ mg cos θ Thus, the equations governing the motion up to the separation point are ˙ N − mg cos θ = −mRθ2 and ¨ mg sin θ = mRθ ˙ (b) Now, multiplying the circumferential equation through by θ we find ˙ ˙¨ mg θ sin θ = mRθθ Integrating once yields 1˙ −g cos θ = Rθ2 + constant 2 ˙ But, θ = 0 when the sphere is at the top of the cylinder where θ = 0. Thus, −g = constant ˙ Solving for θ2 , we find 2g ˙ θ2 = (1 − cos θ) R 4 =⇒ 1˙ − g cos θ = Rθ2 − g 2 =⇒ g 1 d ˙2 d (− cos θ) = R θ dt 2 dt ˙ (c) Substituting the result for θ2 from Part (b) into the radial equation of motion, we have N − mg cos θ = −mR Thus, solving for N yields N = mg (3 cos θ − 2) (d) When the normal reaction force vanishes, the sphere has reached the angle, θs , at which it separates from the cylinder. Setting N = 0 in the equation derived in Part (c), there follows cos θs = Thus, the separation angle is θs = cos−1 2 3 2 3 2g (1 − cos θ) = −2mg (1 − cos θ) R 5 ...
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This note was uploaded on 10/12/2009 for the course AME 301 at USC.

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