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Answer :
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]
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Final answer:
To find the angle between the string and the vertical, use the relationship between centripetal force and the gravitational force acting on the mass. Equate the horizontal component of the tension to the centripetal force requirement, and the vertical component to the gravitational force, then solve for the angle using trigonometric functions.
Explanation:
To calculate the angle between the string and the vertical when a mass is suspended from a string and revolves in a horizontal circle, first recognize that centripetal force is provided by the horizontal component of the tension in the string, while the vertical component balances the weight of the mass.
In this scenario, two forces act on the mass: tension (T) in the string and gravitational force (mg). The tension has two components: Th providing the centripetal force (Fc) necessary for circular motion, and Tv equal to the weight of the mass. Using these forces, the following equations can be formulated:
For horizontal (centripetal) force: Th = T × sin(θ) = m × v2/r
For vertical force balance: Tv = T × cos(θ) = m × g
By dividing these two equations, we eliminate T and can solve for θ:
θ = atan(m × v2 / (r × m × g))
Now plugging in the values (m = 4.10 kg, v = 2.85 m/s, r = 1.69 m, g = 9.8 m/s2), we can determine the angle θ between the string and the vertical.
