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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.)

Short Answer

Expert verified

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

Step by step solution

01

Determination of angle of fishing line

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.

02

Given Data

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

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

03

Apply Newton’s law of motion using the free body diagram

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,

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

04

Find the angle made by the fishing line with vertical

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 \(56^\circ \).

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Most popular questions from this chapter

A jet pilot takes his aircraft in a vertical loop (Fig. 5–38).

(a) If the jet is moving at a speed of840km/hat the lowest point of the loop, determine the minimum radius of the circle so that the centripetal acceleration at the lowest point does not exceed 6.0 g’s. (b) Calculate the 78-kg pilot’s effective weight (the force with which the seat pushes up on him) at the bottom of the circle, and (c) at the top of the circle (assume the same speed).

FIGURE 5-38 Problem 12.

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(a) Find the speed of the bucket.

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