Dynamics_Part33

# Dynamics_Part33 -

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Problem 5: Problem 1. Problem: Three identical spheres of mass m and coefficient of restitution e are initially arranged as shown. All three spheres are attached to the upper wall with chords of the same length. The distance between spheres B and C is negligibly small compared to sphere diameter. Sphere A is released and we wish to determine the value of e based on C , the maximum height reached by Sphere C. (a) Determine the horizontal speed of Sphere A, VA , at the instant when it strikes Sphere B. Express your answer in terms of length and gravitational acceleration, g. (b) Compute the speeds of Sphere A and Sphere B, VA and VB , respectively, immediately after the impact between Spheres A and B. Express your answers in terms of , g and e. (c) Compute the speeds of Sphere B and Sphere C, VB and VC , respectively, immediately after the impact between Spheres B and C. Express your answers in terms of , g and e. (d) If the maximum height reached by Sphere C is C = 8 81 , what is the value of e? n ~ A ................................................................................................................................................................................................................................................... ................................................................................................................................................................................................................................................... ................................................................................................................................................................................................................................................... ................................................................................................................................................................................................................................................... .. .. ................................................................................................................................................................................................................................................. ...... .. ............................................................................................................................................................................................................................................... .. . . . .. . . .. . . . .... ... . . . . . . .. . . . .... . . . . . ... . ... . . . . . . .... . .... . . . . . . . . . . . . . . .... .... . . . . . . . . . . . . . . .... . .... . . . . . . . . . . . . . .... . . . .... . . . . . . . . . .... . . . .... . . . . . . .... .... ........ ....... . . . . . . . .. . . . .. . ... . . ... . .. . . . . .. . . . . .. . . . . . . . . . . . . . . . . 1 . . . . . . . . . . . . . . . 2 . . . . . . . . . . . . . . . . . . . . . .. .. . . . . .. . . . ........ ....... . . . . . ........ ....... g n n ~ ~ B C Solution: To solve this problem, we first use energy conservation to determine the velocity of Sphere A just prior to its impact with Sphere B. Next, we determine the velocity of Sphere B just after the impact. Using this as the initial velocity of Sphere B as it impacts Sphere C, we determine the velocity of Sphere C just after the impact. Finally, using energy conservation, we compute the maximum height to which Sphere C will rise and we infer the value of e. (a) As Sphere A moves from its initial position to the instant just before it impacts Sphere B, total mechanical energy is conserved, i.e., T1 + V1 = T2 + V2 Since Sphere A starts from rest, we know that T1 = 0. Also, setting the origin of our coordinate system at the centers of the spheres when they are in the vertical position, Sphere A's potential energy just prior to impacting Sphere B is V2 = 0. Thus, we have 1 1 2 0 + mg = mVA + 0 2 2 Solving for VA gives VA = g (b) Turning to the impact between Spheres A and B, horizontal momentum is conserved, which tells us that mVA + 0 = mVA + mVB ...
View Full Document

## This note was uploaded on 05/12/2010 for the course AME 301 taught by Professor Shiflett during the Spring '06 term at USC.

Ask a homework question - tutors are online