Question

The 5 lb collar is released from rest at 1 and travels along the smooth guide. Determine its speed when its center reaches point c and the normal force it exerts on the rod at this point. the spring has an unstretched length of 12 in and point c is locatd just before the end of the curved portion of the rod at this point.

Answer 1

a) 12.55ft/s

b) N = 5.392lb

Step-by-step explanation:

Image to question is attached

•Displacement of spring at A

óa = (10+12)-12 = 12in

• displacement of spring at C

óc =
`√(12^2+10^2) - 12 = 3.62`

a)for potential energy of collar at A

(TA)g = Wc ha

= 5(10+12) = 110lb.in

For potential energy of collar at C

(Tc)g = Wc*hc

= 5*10= 50lb.in

For potential energy of spring at A

(TA)s = 1/2 *k*óa²

= (1/2)*(2)(10²) = 100lb.in

Potential energy of collqr at c

(Tc)s = 1/2* 2* 3.62²= 13.12

Kinetoc energy of collar at C =

Vc = (1/2)(Wc/g)Vc²

= (1/2)(5/32.2*12) Vc²

= 0.00647Vc² lb.in

From conservation of mechanical energy

Ea= Ec

(Ta)g + (Ta)s = (Tc)g + (Tc)s + Vc

= 110 +100 = 50+13.1+0.00647Vc²

Vc= 150.68 in/s => 12.55ft/s

b) using free body diagram (attached)

let's find the angle made by spring at point C

We have:

∅ =
`tan^-^1[12/10]`

= 50.19

Therefore, apply force on the horizontal direction, we have:

£fn =
`m_c* a_n`

f_4 sin∅ - N = (Wc/g)(Vc²/12)

= (2*3.62)(sin50.19°) - N = [5/(32.2*12)][12.55²/12]

N = 5.392lb