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Assignment-2-Solutions

# Assignment-2-Solutions - 13 Question 28 A bead slides along...

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13 Question 2–8 A bead slides along a fixed circular helix of radius R and helical inclination angle φ as shown in Fig. P2-8. Knowing that the angle θ measures the position of the bead and is equal to zero when the bead is at the base of the helix, determine the following quantities relative to an observer fixed to the helix: (a) the arclength parameter s as a function of the angle θ , (b) the intrinsic basis { e t , e n , e b } and the curvature of the trajectory as a function of the angle θ , and (c) the position, velocity, and acceleration of the particle in terms of the intrinsic basis { e t , e n , e b } . A O P R θ φ Figure P2-8 Solution to Question 2–8 Let F be a reference frame fixed to the helix. Then, choose the following coor- dinate system fixed in reference frame F : Origin at O E x = Along OA E z = Out of page E y = E z × E x Next, let A be a reference frame that rotates with the projection of the posi- tion of particle into the E x , E y -plane. Corresponding to A , we choose the

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14 Chapter 2. Kinematics following coordinate system to describe the motion of the particle: Origin at O e r = Along O to projection of P into E x , E y plane e z = E z e θ = e z × e r Now, since φ is the angle formed by the helix with the horizontal, we have from the geometry that z = R θ tan φ (2.56) Suppose now that we make the following substitution: α tan φ (2.57) Then the position of the bead can be written as r = R e r + tan φ R θ e z = R e r + α R θ e z (2.58) Furthermore, the angular velocity of reference frame A in reference frame F is given as F ω A = ˙ θ e z (2.59) Then, applying the rate of change transport theorem to r between reference frames A and F , we have F v = F d r dt = A d r dt + F ω A × r (2.60) where A d r dt = α R ˙ θ e z (2.61) F ω A × r = ˙ θ e z × (R e r + α R θ e z ) = R ˙ θ e θ (2.62) Adding Eqs. (2.61) and (2.62), we obtain F v = R ˙ θ e θ + α R ˙ θ e z (2.63) The speed in reference frame F is then given as F v = F v = R ˙ θ 1 + α 2 d dt F s (2.64) Consequently, F ds = R 1 + α 2 d θ (2.65) Integrating both sides of Eq. (2.65), we obtain F s F s 0 ds = θ θ 0 R 1 + α 2 d θ (2.66)
15 We then obtain F s - F s 0 = R 1 + α 2 ( θ - θ 0 ) (2.67) Solving Eq. (2.67) for s , the arclength is given as F s = F s 0 + R 1 + α 2 ( θ - θ 0 ) (2.68) Intrinsic Basis and Curvature of Trajectory The intrinsic basis is obtained as follows. First, the tangent vector e t is given as e t = F v F v (2.69) Substituting the expressions for F v and F v from part (a) into Eq. (2.69), we obtain e t = R ˙ θ e θ + α R ˙ θ e z R ˙ θ 1 + α 2 (2.70) Simplifying this last expression, we obtain e t = e θ + α e z 1 + α 2 (2.71) Next, we have that F d e t dt = κ F v e n (2.72) where F d e t dt = A d e t dt + F ω A × e t (2.73) where A d e t dt = 0 F ω A × e t = ˙ θ e z × e θ + α e z 1 + α 2 = - ˙ θ 1 + α 2 e r (2.74) Therefore, F d e t dt = - ˙ θ 1 + α 2 e r (2.75) The principle unit normal is then given as e n = F d e t /dt F d e t /dt = - e r (2.76) Furthermore, the curvature is given as κ = F d e t /dt F v = 1 R( 1 + α 2 ) (2.77)

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16 Chapter 2. Kinematics Finally, the principle unit bi-normal vector is given as e b = e t × e n = e θ + α e z 1 + α 2 × ( - e r ) = e z - α e θ 1 + α 2 (2.78)
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Assignment-2-Solutions - 13 Question 28 A bead slides along...

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