Question

A 1.6-lb collar is attached to a spring and slides without friction along a circular rod in a vertical plane. The spring has an unstretched length of 6 in. The collar is released from being held at A, and it passes ponit B with a speed of 16.0 ft/s. Determine the spring constant. The circular rod is kept from moving.

Answer 1

Answer:

k = 652 lb/ft

Explanation:

Given :

Weight of the collar = 1.6 lb

The upstretched length of the spring = 6 in

Speed  = 16 ft/s

PA = 8 + 10

     = 18 inch

Let the initial elongation be $\Delta x_i$

∴ $\Delta x_i$ = 18 - 6

         = 12 inch = 1 foot

$PB = \sqrt{13^2+5^2}$

      = 13.925 inch

Final elongation in the spring

$\Delta x_B = 7.928$ inch = 0.66 feet

Applying the conservation of the mechanical energy between A and B is

$K.E_A+P.E_{g,A}+P.E_{sp,A}= K.E_B+P.E_{g,B}+P.E_{sp,B}$

$0+mg_r+\frac{1}{2}k(\Delta x_i)^2=\frac{1}{2}mv_B^2+0+\frac{1}{2}k(\Delta x_B)^2$

$\frac{1}{2}k[(1)^2-(0.66)^2]=\frac{1.6}{2}\times (16)^2-1.6 \times 32 \times \frac{5}{12}$

$0.281 \ k =204.8-21.33$

k = 652 lb/ft