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.

The tangential speed of the mass is 2.85 m/s. Calculate the angle between the string and the vertical.

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) = \frac{mv^2}R`

where
`m=4.10\,\rm kg`,
`v=2.85(\rm m)/(\rm s)`, and
`R` is the radius of the circular path.

As shown in the diagram, we can see that

`\sin(\theta) = \frac Rr \implies R = r\sin(\theta)`

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

`F_(\rm tension) \sin(\theta) = \frac{mv^2}R \\\\ \implies F_(\rm tension) = (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) = (mg)/(\cos(\theta))`

Solve for
`\theta` :

`(mv^2)/(r\sin^2(\theta)) = (mg)/(\cos(\theta)) \\\\ \implies (\sin^2(\theta))/(\cos(\theta)) = (v^2)/(rg) \\\\ \implies (1-\cos^2(\theta))/(\cos(\theta)) = (v^2)/(rg) \\\\ \implies \cos^2(\theta) + (v^2)/(rg) \cos(\theta) - 1 = 0`

Complete the square:

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