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Assignment-3-Solutions - Chapter 3 Kinetics of Particles...

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Chapter 3 Kinetics of Particles Question 3–1 A particle of mass m moves in the vertical plane along a track in the form of a circle as shown in Fig. P3-1. The equation for the track is r = r 0 cos θ Knowing that gravity acts downward and assuming the initial conditions θ (t = 0 ) = 0 and ˙ θ (t = 0 ) = ˙ θ 0 , determine (a) the di ff erential equation of motion for the particle and (b) the force exerted by the track on the particle as a function of θ . r = r 0 cos θ g m O θ Figure P3-1
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62 Chapter 3. Kinetics of Particles Solution to Question 3–1 Kinematics Let F be a reference frame fixed to the track. Then, choose the following coor- dinate system fixed in reference frame F : Origin at point O E x = along OP at t = 0 E z = Out of Page E y = E z × E x Next, let A be a reference frame fixed to the direction OP . Then, choose the following coordinate system fixed in reference frame A : Origin at point O e r = along OP E z = out of page e θ = E z × e r The geometry of the bases { E x , E y , E z } and { e r , e θ , e z } is shown in Fig. 3-1. Using Fig. 3-1, we have that E x = cos θ e r - sin θ e θ (3.1) E y = sin θ e r + cos θ e θ (3.2) e r e θ E x E y θ θ Figure 3-1 Geometry of Coordinate System for Question 3.1 Next, the position of the particle is given in terms of the basis { e r , e θ , e z } as r = r e r = r 0 cos θ e r (3.3) Furthermore, since the angle θ is measured from the fixed horizontal direction, the angular velocity of A in F is given as F ω A = ˙ θ E z (3.4)
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63 Applying the rate of change transport theorem to r between reference frames A and F , we obtain the velocity of the particle in reference frame F as F v = F d r dt = A d r dt + F ω A × r (3.5) Now we have A d r dt = - r 0 ˙ θ sin θ e r (3.6) F ω A × r = ˙ θ E z × r 0 cos θ e r = r 0 ˙ θ cos θ e θ (3.7) Adding the expressions in Eq. (3.6) and Eq. (3.7), we obtain the velocity in refer- ence frame F as F v = - r 0 ˙ θ sin θ e r + r 0 ˙ θ cos θ e θ (3.8) Re-writing Eq. (3.8), we obtain F v = r 0 ˙ θ ( - sin θ e r + cos θ e θ ) (3.9) The speed in reference frame F is then given as F v = r 0 ˙ θ (3.10) Dividing F v by F v , we obtain the tangent vector as e t = - sin θ e r + cos θ e θ (3.11) Next, from the first of the Serret-Frenet formulae, we have that F d e t dt = κ F v e n (3.12) Using the rate of change transport theorem, we have that F d e t dt = A d e t dt + F ω A × e t (3.13) where A d e t dt = - ˙ θ cos θ e r - ˙ θ sin θ e θ (3.14) F ω A × e t = ˙ θ E z × ( - sin θ e r + cos θ e θ ) = - ˙ θ cos θ e r - ˙ θ sin θ e θ (3.15) Adding the two expressions in Eq. (3.15) and Eq. (3.15), we have that F d e t dt = - 2 ˙ θ cos θ e r - 2 ˙ θ sin θ e θ (3.16)
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64 Chapter 3. Kinetics of Particles Then the principle unit normal in reference frame F is given as e n = F d e t /dt F d e t /dt = - cos θ e r - sin θ e θ (3.17) Also, the curvature is given as κ = 1 F v F d e t /dt F v = 2 ˙ θ r 0 ˙ θ = 2 r 0 (3.18) Finally, di ff erentiating Eq. (3.10), we obtain d dt F v = r 0 ¨ θ (3.19) The acceleration of the particle in reference frame F is then given as F a = d dt F v + κ F v 2 e n = r 0 ¨ θ e t + 2 r 0 r 2 0 ˙ θ 2 e n (3.20) which simplifies to F a = r 0 ¨ θ e t + 2 r 0 ˙ θ 2 e n (3.21) Kinetics Next, in order to obtain the di ff
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