Solving a Midterm Physics Problem: Masses A & B with Springs

[DCholic]

Homework Statement

Masses A and B, each having a mass of 32.2 slugs, are constrained to move in frictionless slots. They are connected by a rigid, massless rod of length L = 2 ft. Mass B is connected to two massless linear springs, each having a spring constant k = 60 lbF / ft. The springs are unstretched when the connecting rod to masses A and B is vertical.

The accompanying schematic:
https://drive.google.com/file/d/0B2-REaK_0ruSZ3cwV2dZS1VuNVE/view?usp=sharing

If B is given a slight nudge, what are the velocities of B and A when A descends a distance of D = 3 inches?

Homework Equations

W_1->2 = ΔKE + ΔPE_G + ΔPE_E

The Attempt at a Solution

Masses: m_A = m_B = m
Take upward to be positive y and right as positive x.

At position 1:
Velocity of A: v_A1 = 0
Velocity of B: v_B1 = 0
Height of A: h_A1 = L
Height of B: h_B1 = 0
Displacement of left spring: x_L1 = 0
Displacement of right spring: x_R1 = 0

Finding the displacement of each spring:
(L - D)² + x²= L²
x = 11.62 in. = 0.9682 ft -> symmetric for both springs

At position 2:
Velocity of A: v_A2 = ?
Velocity of B: v_B2 = ?
Height of A: h_A2 = L - D
Height of B: h_B2 = 0
Displacement of left spring: x_L2 = x
Displacement of right spring: x_R2 = x

Work:
W_1->2 = 0 because there are no non-conservative forces acting on the system

Change in kinetic energy:
ΔKE = (1/2) * m_A * (v_A2² - v_A1²) + (1/2) * m_B * (v_B2² - v_B1²)
ΔKE = (1/2) * m * (v_A2² + v_B2²)

Change in gravitational potential energy: g = 32.174 ft/s²
ΔPE_G = m_A * g * (h_A2 - h_A1) + m_B * g * (h_B2 - h_B1)
ΔPE_G = m * g * ( - D) = -259.001 slug⋅ft²/s² = -259.001 ft⋅lbF

Change in elastic potential energy:
ΔPE_E = (1/2) * k * (x_L2² - x_L1²) + (1/2) * k * (x_R2² - x_R1²)
ΔPE_E = (1/2) * k * ( 2 * x²)
ΔPE_E = k * x² = 56.245 ft⋅lbF

Relating v_A2 and v_B2:
Take the origin at B's initial position.
x ≡ horizontal distance from origin to B
y ≡ vertical distance from origin to A

x² + y² = L² = constant
2 * x * (dx/dt) + 2 * y * (dy/dt) = 0
2 * x * v_B + 2 * y * v_A = 0
x * v_B + y * v_A = 0

At the instant when A has moved down 3 inches:
v_B2 = [ - (L - D) / x] * v_A2 = -1.807 * v_A2

Combine all equations:
W_1->2 = ΔKE + ΔPE_G + ΔPE_E
0 = (1/2) * m * [v_A2² + ( -1.807 * v_A2)²] - 259.001 + 56.245
v_A2 = 1.718 ft/s down
v_B2 = 3.105 ft/s right

This was a midterm question. Physics enthusiasts, please give feedback my thought process. Much appreciated.

 

[TSny]

Welcome to PF!

Nicely organized. Your work looks correct to me.