University of California, Berkeley Department of Mechanical Engineering ME 104, Fall Midterm Exam 1 Solutions

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1 University of California, Berkeley Department of Mechanical Engineering ME 104, Fall 2013 Midterm Exam 1 Solutions 1. (20 points) (a) For a particle undergoing a rectilinear motion, the position, velocity, and acceleration vectors are given by r = x i, v = ẋ i = v i, a = ẍ i = a i. (1) Show that a = d ( ) 1 dx 2 v2. (2) d dx ( ) 1 2 v2 = 1 2 (2v)dv dx = v dv dx = dx dv dt dx = dv dt = a (b) Consider a particle P of mass m travelling in a horizontal plane. Let its velocity vector be v = ṙ = ẋ i + ẏ j = v x i + v y j. (3) Suppose that v 2 x and v 2 y are specified by v 2 x = C kx2 0, v 2 y = C ky2 0, (4) where k is a positive constant, and C 1 and C 2 are constants. Calculate a x and a y. Using (2), a x = d dx [ 1 (C 1 12 )] 2 kx2 = kx 2, a y = d [ 1 (C 2 12 )] dy 2 ky2 = ky 2 (c) Show that the force acting on P is centripetal. The force acting on P can be obtained using Newton s second law: F = ma F x i + F y j = ma x i + ma y j. 1

2 Taking the inner product with i and j respectively, and using the results from Part (b), Then, F x = ma x = mkx 2, F y = ma y = mky 2. F = F x i + F y j = mk (x i + y j) = mk 2 2 e r. The direction of the force is e r. The force acting on P is centripetal. (d) Obtain the differential equations for the x and y coordinates of P, and provide a simple solution of either one of them. From the acceleration components a x and a y, we have ẍ + k 2 x = 0, ÿ + k 2 y = 0. Both equations represent simple harmonic motion. We can suppose that the solution has the form x = A cos(ωt) + B sin(ωt). Differentiate once to get Differentiate again to get Substituting into the differential equation, ẋ = ωa sin(ωt) + ωb cos(ωt). ẍ = ω 2 [A cos(ωt) + B sin(ωt)]. ω 2 [A cos(ωt) + B sin(ωt)] + k [A cos(ωt) + B sin(ωt)] = 0 2 ( [A cos(ωt) + B sin(ωt)] ω 2 + k ) = 0 2 So ω = k/2. The solution becomes x = A cos ( ) ( ) k k 2 t + B sin 2 t We can choose the initial conditions to solve for A and B and complete the solution. 2

3 2. (30 points) Let OABC be a rigid plate which is rotating at constant angular velocity ω = ω k around a vertical axis OZ (ω = θ = const.). Introduce a corotational basis {e r, e θ } on the plate, as indicated in Fig. 1. Suppose that a rigid rod OAD is welded to the plate and that MN is a rigid guide (OMN is a right angle.). Let P be a pin of mass m that can slide on the rod and inside the guide. N B v D 2b b r P C A e θ j e r θ M O i Figure 1: Problem 2 Figure 1. (a) Write down the relationship between the corotational basis vectors e r, e θ and the fixed basis vectors i, j. (Solution) (4 points) 3

4 This can either be written in matrix form: ( ) ( ) ( ) er cos θ sin θ i =, sin θ cos θ j or e r = cos θ i + sin θ j e θ = sin θ i + cos θ j. e θ (b) Show that ė r = θ e θ = ω e r, ė θ = θ e r = ω e θ. (5) Differentiating the cylindrical basis with respect to t, ė r = de r dt = θ ( sin θ) i + θ cos θ j = θ e θ, Also, ė θ = de θ dt = θ cos θ i + θ ( sin θ) j = θ e r ω e r = θ k e r = θ e θ ω e θ = θ k e θ = θ e r (c) Express the velocity and acceleration of P on both bases, given that θ = const. Hence deduce that ṙ = ẏ j e r, r θ = ẏ j e θ. (6) We note that in the motion of the pin, v x = ẋ = 0. If r = r(t) e r (t) = x i + y j, and with θ = const, v = ṙ = ẏ j = v y j = ṙ e r + r ė r = ṙ e r + r θ e θ = v r e r + v θ e θ 4

5 a = v = ÿ j = a y j = r e r + ṙė r + ṙ θe θ + r θe θ = r e r + ṙ θ e θ + ṙ θ e θ r θ 2 e r = ( r r θ 2 ) e r + (2ṙ θ) e θ = a r e r + a θ e θ. We can take the inner product of the velocity with e r and e θ to get the equations in (6): v e r = ṙ = ẏ j e r, v e θ = r θ = ẏ j e θ (d) Suppose that at θ = 30, r = 0.05 m, ṙ = 0.2 m/s, r = m/s 2. (7) (i) Calculate θ and ẏ, and check that ẏ 2 = ṙ 2 + r 2 θ2. (ii) Find ÿ. (iii) Calculate the magnitude of the total force F acting on P if the mass of P is m = 0.5 kg. (Solution) (7 points) The results from parts (a) and (c) gives us the component equations (i) Dividing the equations, Also, ṙ = ẏ j e r = ẏ sin θ, and r θ = ẏ j e θ = ẏ cos θ. ṙ r θ = tan θ ṙ θ = r tan θ 0.2 m/s = 0.05 m tan 30 1/5 = 1/20 1/ 3 rad/s = 4 3 rad/s = 6.93 rad/s. ṙ = ẏ sin θ ẏ = ṙ sin θ 0.2 m/s = sin 30 = 0.4 m/s. 5

6 Then ẏ 2 = ṙ 2 + r 2 θ2 (0.4 m/s) 2 = (0.2 m/s) 2 + (0.05 m) m 2 /s 2 = 0.04 m 2 /s m 2 /s 2 = 0.16 m 2 /s 2. (ii) You are correct if you used the following methods to find ÿ. ( 4 ) 2 3 rad/s (1) ÿ j j = a j = ( r r θ 2 ) e r j + (2ṙ θ) e θ j or (2) ÿ j e r = a e r = r r θ 2 or (3) ÿ j e θ = a e θ = 2ṙ θ It is given in the problem that θ = const. With the introduction of a value for r, the problem actually becomes over-determined. This means a value is specified which can actually be calculated, and if we don t choose to specify it correctly, we have a contradiction. As a result, ÿ does not have a unique solution. For the sake of completeness, the actual values of r and ÿ are solved below. ÿ j e θ = 2ṙ θ 2ṙ θ ÿ = cos θ = = 3.2 m/s 2 cos 30 ÿ j e r = r r θ 2 r = ÿ sin θ + r θ 2 = 3.2 sin (4 3) 2 = 4 m/s 2 (iii) The magnitude of the force is F = ma = m a = m ÿ j = mÿ = 0.5 kg 3.2 m/s 2 = 1.6 N. (e) Let η represent the distance along the diagonal OB of the plate. Calculate the velocity distribution v(η) along OB and sketch it. We can set up a new basis {e 1, e 2 } by rotating the basis {e r, e θ } at an angle φ = tan 1 (1/2) = such that e 1 is along the diagonal of the plate and e 2 is perpendicular to it. See Figure 2. The unit vector e 1 along the diagonal of the rigid plate is e 1 = cos φ e r + sin φ e θ = 1 5 (2 e r + e θ ). 6

7 ω B 2b b C η A e θ e 1 e 2 ϕ ϕ θ e r O i Figure 2: Rotating plate The unit vector perpendicular to the diagonal is Also, e 1 e 2 = k. Their time derivatives are Along OB, The velocity is e 2 = sin φ e r + cos φ e θ = 1 5 ( e r + 2 e θ ). ė 1 = ω e 1 = θ e 2, ė 2 = ω e 2 = θ e 1. r(η) = η e 1 = η (cos φ e r + sin φ e θ ) = η 5 (2 e r + e θ ). v = ṙ = η ė 1 = η ω e 1 = ω r = θ k η e 1 = θη k e 1 = η θ e 2 = η θ ( sin φ e r + cos φ e θ ). A sketch of the velocity is shown in Figure 3. 7

8 (f) Also calculate the acceleration along the diagonal OB and indicate its variation on a sketch ( θ = const.). (Solution) (4 points) The acceleration is a = v = η θ ė 2 = η θ ω e 2 = ω (ω r) = θ k (η θ k e 1 ) = η θ 2 e 1 = θ 2 r(η), which is centripetal. A sketch is shown in Figure 3. v(η) B B 2b b 2b b C A C a(η) A e θ e 1 e r O O Figure 3: Velocity and acceleration 8

Exam 1 September 11, 2013

Exam 1 September 11, 2013 Exam 1 Instructions: You have 60 minutes to complete this exam. This is a closed-book, closed-notes exam. You are allowed to use an approved calculator during the exam. Usage of mobile phones and other