Question

A mass of 4.10 kg is suspended from a 1.69 m long string. It revolves in a horizontal circle as shown in the figure.

Answer 1

The horizontal component of the tension in the string is a centripetal force, so by Newton's second law we have

• net horizontal force

$F_{\rm tension} \sin(\theta) = \dfrac{mv^2}R$

where $m=4.10\,\rm kg$, $v=2.85\frac{\rm m}{\rm s}$, and $R$ is the radius of the circular path.

As shown in the diagram, we can see that

$\sin(\theta) = \dfrac Rr \implies R = r\sin(\theta)$

where $r=1.69\,\rm m$, so that

$F_{\rm tension} \sin(\theta) = \dfrac{mv^2}R \\\\ \implies F_{\rm tension} = \dfrac{mv^2}{r\sin^2(\theta)}$

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

$F_{\rm \tension} \cos(\theta) - mg = 0 \\\\ \implies F_{\rm tension} = \dfrac{mg}{\cos(\theta)}$

Solve for $\theta$ :

$\dfrac{mv^2}{r\sin^2(\theta)} = \dfrac{mg}{\cos(\theta)} \\\\ \implies \dfrac{\sin^2(\theta)}{\cos(\theta)} = \dfrac{v^2}{rg} \\\\ \implies \dfrac{1-\cos^2(\theta)}{\cos(\theta)} = \dfrac{v^2}{rg} \\\\ \implies \cos^2(\theta) + \dfrac{v^2}{rg} \cos(\theta) - 1 = 0$

Complete the square:

$\cos^2(\theta) + \dfrac{v^2}{rg} \cos(\theta) + \dfrac{v^4}{4r^2g^2} = 1 + \dfrac{v^4}{4r^2g^2} \\\\ \implies \left(\cos(\theta) + \dfrac{v^2}{2rg}\right)^2 = 1 + \dfrac{v^4}{4r^2g^2} \\\\ \implies \cos(\theta) + \dfrac{v^2}{2rg} = \pm \sqrt{1 + \dfrac{v^4}{4r^2g^2}} \\\\ \implies \cos(\theta) = -\dfrac{v^2}{2rg} \pm \sqrt{1 + \dfrac{v^4}{4r^2g^2}}$

Plugging in the known quantities, we end up with

$\cos(\theta) \approx 0.784 \text{ or } \cos(\theta) \approx -1.27$

The second case has no real solution, since $-1\le\cos(\theta)\le1$ for all $\theta$. This leaves us with

$\cos(\theta) \approx 0.784 \implies \theta \approx \cos^{-1}(0.784) \approx \boxed{38.3^\circ}$