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Unformatted text preview: 64 Chapter 3. Kinetics of Particles Question 35 A collar of mass m is constrained to move along a frictionless track in the form of a logarithmic spiral as shown in Fig. P35. The equation for the spiral is given as r = r e a where r and a are constants and is the angle as shown in the figure. Assuming that gravity acts downward, determine the differential equation of motion in terms of the angle using (a) Newtons 2 nd law and (b) the workenergy theorem for a particle. r = r e a g m O Figure P35 Solution to Question 35 Kinematics Let F be a reference frame fixed to the track. Then, choose the following coordinate system fixed in reference frame F : Origin at O E x = To the Right E z = Out of Page E y = E z E x Next, let A be a reference frame fixed to the direction of Om . Then, we choose the following coordinate system fixed in reference frame F : Origin at O e r = Along Om E z = Out of Page e = E z e r 65 The relationship between the bases { E x , E y , E z } and { e r , e , e z } is given as e r = cos E x + sin E y (3.143) e =  sin E x + cos E y (3.144) The position of the particle is then given as r = r e r = r e a e r (3.145) Furthermore, the angular velocity of reference frame A in reference frame F is given as F A = E z (3.146) Applying the rate of change transport theorem 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 A r (3.147) where A d r dt = r e r =  ar e a e r (3.148) F A r = E z r e r = E z r e a e r = r e a e (3.149) Adding the expressions in Eq. (3.148) and Eq. (3.149), we obtain the velocity of the particle in reference frame F as F v =  ar e a e r + r e a e (3.150) Simplifying Eq. (3.147), we obtain F v as F v = r e a [ a e r + e ] (3.151) Now we need the acceleration of the collar in reference frame F . For this problem it is most convenient to obtain F a in terms of an intrinsic basis as viewed by an observer fixed to the track. First, the tangent vector is given as e t = F v bardbl F v bardbl = F v bardbl F v bardbl (3.152) where...
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This note was uploaded on 04/03/2012 for the course AERO 310 taught by Professor Chakravorty during the Spring '07 term at Texas A&M.
 Spring '07
 Chakravorty
 Dynamics

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