While fishing, you get bored and start to swing a sinker weight around in a circle below you on a 0.25 m piece of fishing line. The weight makes a complete circle every 0.75 s. What is the angle that the fishing line makes with the vertical? (Hint: See Fig 5-10.)

Using Newton’s law, the equation of force along a fishing line can be calculated. The horizontal component of the force along a fishing line is equal to the centripetal force.

The distance between the sniker and the circle is \(L = 0.25\;m\).

The time taken to complete a circle is \(T = 0.75\;s\).

The figure below represents the free body diagram, with representation of force.

The radius of the circle of motion is \(r = L\sin \theta \).

Apply Newton’s law in the vertical direction is given as,

\(\begin{aligned}\sum {{F_y}} &= 0\\{F_T}\cos \theta - mg &= 0\\{F_T} &= \frac{{mg}}{{\cos \theta }}\end{aligned}\)

Apply Newton’s law for radial forces is given as,

\(\begin{aligned}\sum {{F_R}} &= m{a_R}\\{F_T}\sin \theta &= m\frac{{{v^2}}}{r}\end{aligned}\)

Simplify the equation to find the angle of fishing line with vertical.

\(\begin{aligned}{F_T}\sin \theta &= m\frac{{{v^2}}}{r}\\\left( {\frac{{mg}}{{\cos \theta }}} \right)\sin \theta &= \frac{m}{r}{\left( {\frac{{2\pi r}}{T}} \right)^2}\\\left( {\frac{{mg}}{{\cos \theta }}} \right)\sin \theta &= \frac{{4{\pi ^2}m\left( {L\sin \theta } \right)}}{{{T^2}}}\\\cos \theta &= \frac{{g{T^2}}}{{4{\pi ^2}L}}\end{aligned}\)

Substitute the values in the above equation,

\(\begin{aligned}\theta &= {\cos ^{ - 1}}\frac{{\left( {9.8\;{\rm{m/}}{{\rm{s}}^2}} \right){{\left( {0.75\;{\rm{s}}} \right)}^2}}}{{4{\pi ^2}\left( {0.25\;{\rm{m}}} \right)}}\\ &= 56^\circ \end{aligned}\)

Thus, the angle of fishing line with the vertical line is \(56^\circ \).