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Midterm examination: Dynamics Hiroki Okubo June 1 2011 y θ A path ρ r eθ er A r θ Figure 2: Sliding block. x O Figure 1: Polar coordinates. (c) Differentiate the position vector with respect to the time and show the vector expression for the velocity. (5) Solution. We differentiate the position vector. 1. Figure 1 shows the polar coordinates r and θ which locate a particle traveling on a curved path. Unit vectors er and eθ are established in the positive r- and θ-directions, respectively. dr dt = (a) Determine the position vector r to the particle at A. (5) Solution. The position vector is r = rer . (b) Find the time derivatives of er and eθ using er and eθ , respectively. (5) Solution. Unit vectors i and j are attached to the xy axes. The er and eθ can be described as er eθ = cos θi + sin θj = − sin θi + cos θj deθ dt = θ̇(− sin θi + cos θj) = θ̇eθ = θ̇(cos θi + sin θj) = −θ̇er der dt ṙer + rθ̇eθ ṙer + r (5) (d) Determine the acceleration. (5) Solution. We differentiate the velocity. d2 r dt2 = = der deθ + ṙ θ̇eθ + rθ̈eθ + rθ̇ dt dt (r̈ − rθ̇2 )er + (rθ̈ + 2ṙθ̇)eθ (6) r̈er + ṙ 2. Determine the maximum speed v which the sliding block may have as it passes point A without losing contact with the surface as shown in Fig. 2. The radius of curvature at A is ρ. (10) Solution. The equation of motion in the normal direction is (1) (2) The time derivatives of them can be calculated by der dt = (3) mg − N = m v2 ρ (7) where N is the reaction force from the surface. The force is zero, N = 0, when the block loses (4) 1 A B where m is the mass, N is the reaction force from the slide and θ = 30◦ . Eliminating N from Eqs. (12) and (13) yields a 30 9.81 a = g tan θ = √ = 5.66 m/s2 3 Figure 3: Sliding collar. 5. The acceleration of a particle, a, is assumed to be written a = −C1 − C2 v 2 where C1 and C2 are constants, and v is the velocity. If the particle has an initial velocity v0 , derive an expression for the distance D required for it to a stop. (10) Solution. Integration of the acceleration gives the distance contact with the surface. Substituting N = 0 into Eq. (7) gives v= √ gρ (8) √ If the speed at A were less than gρ, an upward normal force exerted by the surface on the block would exist. In order for the block to have a speed at A which is greater than √ gρ, some type of constraint would have to be introduced to provide additional downward force. dv dt dv v dt 3. A particle moves in a circular path of 0.4-m radius. Calculate the magnitude a of the acceleration of the particle if its speed is 0.6 m/s but is increasing at the rate of 1.2 m/s each second. (10) Solution. With the constant velocity given, we can compute the acceleration from a= 0.62 = 0.9 m/s2 0.4 Z where the magnitude of a is p a = 0.92 + 1.22 = 1.5 m/s2 = = 0 ma −C1 − C2 v 2 = −(C1 + C2 v 2 ) ds = ds = D = D = (9) ds dt vdv 1 2 2 C1 1 + C C1 v Z 0 1 vdv − C2 2 C1 v0 1 + C v 1 0 C2 2 1 ln | 1 + v | − 2C2 C1 v0 − 1 C2 2 ln | 1 + v | 2C2 C1 0 (15) where s is the displacement of the particle. (10) (11) 4. The collar A is free to slide along the smooth shaft B mounted in the frame in Fig. 3. The plane of the frame is vertical. Determine the horizontal acceleration a of the frame necessary to maintain the collar in a fixed position on the shaft. (10) Solution. The equations of motion of the collar give −mg + N cos θ N sin θ = 0 With the unit vectors en and et , the total acceleration becomes a = 0.9en + 1.2et m/s2 (14) (12) (13) 2