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Unformatted text preview: 140 Chapter 4. Kinetics of a System of Particles Question 43 A block of mass m is dropped from a height h above a plate of mass M as shown in Fig. P4-3. The plate is supported by three linear springs, each with spring constant K , and is initially in static equilibrium. Assuming that the compression of the springs due to the weight of the plate is negligible, that the impact is perfectly inelastic, that the block strikes the vertical center of the plate, and that gravity acts downward, determine (a) the velocity of the block and plate immediately after impact and (b) the maximum compression, x max , attained by the springs after impact. g m h K K K M Figure P4-2 Solution to Question 43 Kinematics Let F be a fixed reference frame. Then, choose the following coordinate system fixed in reference frame F : Origin at m at t = E x = Down E z = Out of Page E y = E z E x Then the velocity of the block and plate in reference frame F are given, respectivley, as F v 1 = v 1 E x (4.91) F v 2 = v 2 E x (4.92) The velocity of the center of mass of the block-plate system in reference frame F is then given as F v = m F v 1 + M F v 2 m + M == mv 1 + Mv 2 m + M E x (4.93) 141 Kinetics Phase 1: During Descent of Block The only force acting on the block during its descent is that of gravity. Since gravity is conservative, we know that energy must be conserved. Consequently, F E(t ) = F E(t 1 ) (4.94) where t and t 1 are the times of the start and end of the descent. Now we have that F E = F T + F U (4.95) Since the block is dropped from rest, we have that F T(t ) = (4.96) Next, the kinetic energy of the block at the end of the descent is given as F T(t ) = 1 2 m F v 1 (t 1 ) 1 2 m F v 1 (t 1 ) = 1 2 mv 2 1 (t 1 ) (4.97) Also, the initial potential energy is given as F U(t ) = - m g r 1 (t ) (4.98) where r 1 (t ) is the position of the block at time t = t . Now since the origin of the co- ordinate system is located at the block at t = 0, we know that r (t ) = . Consequently, F U(t ) = (4.99) Finally, the potential energy at the end of the descent is given as F U(t 1 ) = - m g r 1 (t 1 ) (4.100) Now since the block is dropped from a height h above the plate, we have that...
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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