The horizontal component of the tension in the string is a centripetal force, so by Newton's second law we have
• net horizontal force
[tex]F_{\rm tension} \sin(\theta) = \dfrac{mv^2}R[/tex]
where [tex]m=4.10\,\rm kg[/tex], [tex]v=2.85\frac{\rm m}{\rm s}[/tex], and [tex]R[/tex] is the radius of the circular path.
As shown in the diagram, we can see that
[tex]\sin(\theta) = \dfrac Rr \implies R = r\sin(\theta)[/tex]
where [tex]r=1.69\,\rm m[/tex], so that
[tex]F_{\rm tension} \sin(\theta) = \dfrac{mv^2}R \\\\ \implies F_{\rm tension} = \dfrac{mv^2}{r\sin^2(\theta)}[/tex]
The vertical component of the tension counters the weight of the mass and keeps it in the same plane, so that by Newton's second law we have
• net vertical force
[tex]F_{\rm \tension} \cos(\theta) - mg = 0 \\\\ \implies F_{\rm tension} = \dfrac{mg}{\cos(\theta)}[/tex]
Solve for [tex]\theta[/tex] :
[tex]\dfrac{mv^2}{r\sin^2(\theta)} = \dfrac{mg}{\cos(\theta)} \\\\ \implies \dfrac{\sin^2(\theta)}{\cos(\theta)} = \dfrac{v^2}{rg} \\\\ \implies \dfrac{1-\cos^2(\theta)}{\cos(\theta)} = \dfrac{v^2}{rg} \\\\ \implies \cos^2(\theta) + \dfrac{v^2}{rg} \cos(\theta) - 1 = 0[/tex]
Complete the square:
[tex]\cos^2(\theta) + \dfrac{v^2}{rg} \cos(\theta) + \dfrac{v^4}{4r^2g^2} = 1 + \dfrac{v^4}{4r^2g^2} \\\\ \implies \left(\cos(\theta) + \dfrac{v^2}{2rg}\right)^2 = 1 + \dfrac{v^4}{4r^2g^2} \\\\ \implies \cos(\theta) + \dfrac{v^2}{2rg} = \pm \sqrt{1 + \dfrac{v^4}{4r^2g^2}} \\\\ \implies \cos(\theta) = -\dfrac{v^2}{2rg} \pm \sqrt{1 + \dfrac{v^4}{4r^2g^2}}[/tex]
Plugging in the known quantities, we end up with
[tex]\cos(\theta) \approx 0.784 \text{ or } \cos(\theta) \approx -1.27[/tex]
The second case has no real solution, since [tex]-1\le\cos(\theta)\le1[/tex] for all [tex]\theta[/tex]. This leaves us with
[tex]\cos(\theta) \approx 0.784 \implies \theta \approx \cos^{-1}(0.784) \approx \boxed{38.3^\circ}[/tex]
