This preview shows pages 1–4. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: 152 Chapter 4. Kinetics of a System of Particles Question 4–17 A dumbbell consists of two particles A and B each of mass m connected by a rigid massless rod of length 2 l . Each end of the dumbbell slides without friction along a fixed circular track of radius R as shown in Fig. P417. Knowing that θ is the angle from the vertical to the center of the rod and that gravity acts downward, determine the differential equation of motion for the dumbbell. 2 l g a m m A B C O R θ Figure P412 Solution to Question 4–17 This problem can be solved using many different approaches. For the purposes of this solution, we will show the following three two most prominent of these approaches: (1) using point O as the reference point or (2) using the center of mass as the reference point. In order to apply all of these approaches, we will need to compute the following kinematic quantities: (a) the acceleration of each particle, the angular momentum of the system relative to point O , and (c) the angular momentum of the system relative to the center of mass. 153 Kinematics First, let F be a fixed reference frame. Then, choose the following coordinate system fixed in reference frame F : Origin at Point O E x = Along OC When θ = E z = Out of Page E y = E z × E x Next, let A be a reference frame that rotates with the dumbbell. Then, choose the following coordinate system fixed in reference frame A : Origin at Point O e r = Along OC e z = 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. 46 from which we have that e r e θ E x E y E z , e z θ θ Figure 46 Geometry of Bases { E x , E y , E z } and { e r , e θ , e z } for 4–17. E x = cos θ e r sin θ e θ (4.191) E y =  sin θ e r + cos θ e θ (4.192) It is seen that the angular velocity of reference frame A in reference frame F is given as F ω A = ˙ θ e z (4.193) Furthermore, the position of each particle is then given in terms of the basis { e r , e θ , e z } as r A = a e r l e θ (4.194) r B = a e r + l e θ (4.195) 154 Chapter 4. Kinetics of a System of Particles The velocity of each particle can then be obtained using the transport theorem as F v A = A d r A dt + F ω A × r A (4.196) F v B = A d r B dt + F ω A × r B (4.197) Now since a and l are constant, we have that A d r A dt = (4.198) A d r B dt = (4.199) Furthermore, F ω A × r A = ˙ θ e z × (a e r l e θ ) = l ˙ θ e r + a ˙ θ e θ (4.200) F ω A × r B = ˙ θ e z × (a e r + l e θ ) =  l ˙ θ e r + a ˙ θ e θ (4.201) The velocity of each particle is then given as F v A = l ˙ θ e r + a ˙ θ e θ (4.202) F v A =  l ˙ θ e r + a ˙ θ e θ (4.203) The acceleration of each particle is then obtained from the transport theorem as F a A = A d dt parenleftBig F v A parenrightBig...
View
Full
Document
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

Click to edit the document details