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

k = 652 lb/ft

Step-by-step 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

`$PB = √(13^2+5^2)$`

= 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) $`

`$0+mg_r+(1)/(2)k(\Delta x_i)^2=(1)/(2)mv_B^2+0+(1)/(2)k(\Delta x_B)^2$`

`$(1)/(2)k[(1)^2-(0.66)^2]=(1.6)/(2)* (16)^2-1.6 * 32 * (5)/(12)$`

`0.281 \ k =204.8-21.33`

k = 652 lb/ft