0.45 kg collar and spring on circular rod

Geometry (read from the figure): radius $r=0.127$ m, centre O. Collar released at A (right end of the horizontal diameter), B is the lowest point. The spring is fixed at C, $0.178$ m to the left of the circle, on the horizontal line through O. Take O as origin: A $=(0.127,0)$, B $=(0,-0.127)$, C $=(-0.305,0)$. Undeformed length $0.127$ m, $k=146$ N/m, $m=0.45$ kg.

Spring lengths

$$\begin{aligned} L_A&=0.127+0.305=0.432\ \text{m}\ \Rightarrow\ x_A=0.432-0.127=0.305\ \text{m}\\ L_B&=\sqrt{0.305^{2}+0.127^{2}}=0.3304\ \text{m}\ \Rightarrow\ x_B=0.2034\ \text{m} \end{aligned}$$

Speed at B (conservation of energy, datum at B)

$$mg\,r+\tfrac12kx_A^{2}=\tfrac12mv_B^{2}+\tfrac12kx_B^{2}$$ $$0.45(9.81)(0.127)+\tfrac12(146)(0.305)^{2}=\tfrac12(0.45)v_B^{2}+\tfrac12(146)(0.2034)^{2}$$ $$0.5607+6.7908=0.225v_B^{2}+3.0200\ \Rightarrow\ v_B^{2}=19.25,\ \ v_B=4.388\ \text{m/s}$$

Normal force at B

Spring force $F_s=kx_B=146(0.2034)=29.69$ N, directed from B towards C. Its vertical (upward) component is

$$F_{s,y}=29.69\cdot\frac{0.127}{0.3304}=11.41\ \text{N}$$

The horizontal component is tangential, so it does not enter the normal equation. Newton's second law toward the centre O (upward) at B:

$$N+F_{s,y}-mg=\frac{mv_B^{2}}{r}$$ $$N=\frac{0.45(19.25)}{0.127}+0.45(9.81)-11.41=68.2+4.41-11.41=61.2\ \text{N}$$

Answer: $v_B\approx4.39$ m/s; normal force $N\approx61.2$ N acting upward, towards O.