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Calculating the acceleration of a fishing reel

A deep-sea fisherman hooks a big fish that swims away from the boat pulling the fishing line from his fishing reel. The whole system is initially at rest and the fishing line unwinds from the reel at a radius of 4.50 cm from its axis of rotation. The reel is given an angular acceleration of 110 rad/s 2 size 12{"110""rad/s" rSup { size 8{2} } } {} for 2.00 s as seen in [link] .

(a) What is the final angular velocity of the reel?

(b) At what speed is fishing line leaving the reel after 2.00 s elapses?

(c) How many revolutions does the reel make?

(d) How many meters of fishing line come off the reel in this time?

Strategy

In each part of this example, the strategy is the same as it was for solving problems in linear kinematics. In particular, known values are identified and a relationship is then sought that can be used to solve for the unknown.

Solution for (a)

Here α size 12{α} {} and t size 12{α} {} are given and ω size 12{ω} {} needs to be determined. The most straightforward equation to use is ω = ω 0 + αt size 12{ω=ω rSub { size 8{0} } +αt} {} because the unknown is already on one side and all other terms are known. That equation states that

ω = ω 0 + αt . size 12{ω=ω rSub { size 8{0} } +αt"."} {}

We are also given that ω 0 = 0 size 12{ω rSub { size 8{0} } =0} {} (it starts from rest), so that

ω = 0 + 110 rad/s 2 2 . 00 s = 220 rad/s . size 12{ω=0+ left ("110"" rad/s" rSup { size 8{2} } right ) left (2 "." "00"" s" right )="220 rad/s."} {}

Solution for (b)

Now that ω size 12{ω} {} is known, the speed v size 12{v} {} can most easily be found using the relationship

v = , size 12{v=rω","} {}

where the radius r size 12{α} {} of the reel is given to be 4.50 cm; thus,

v = 0 . 0450 m 220 rad/s = 9 . 90 m/s. size 12{v= left (0 "." "0450"" m" right ) left ("220"" rad/s" right )=9 "." "90"" m/s."} {}

Note again that radians must always be used in any calculation relating linear and angular quantities. Also, because radians are dimensionless, we have m × rad = m size 12{m times "rad"=m} {} .

Solution for (c)

Here, we are asked to find the number of revolutions. Because 1 rev = 2π rad size 12{1" rev"=2π" rad"} {} , we can find the number of revolutions by finding θ size 12{θ} {} in radians. We are given α size 12{α} {} and t size 12{t} {} , and we know ω 0 size 12{ω rSub { size 8{ {} rSub { size 6{0} } } } } {} is zero, so that θ size 12{θ} {} can be obtained using θ = ω 0 t + 1 2 αt 2 size 12{θ=ω rSub { size 8{0} } t+ { {1} over {2} } αt rSup { size 8{2} } } {} .

θ = ω 0 t + 1 2 αt 2 = 0 + 0.500 110 rad/s 2 2.00 s 2 = 220 rad . alignl { stack { size 12{θ=ω rSub { size 8{0} } t+ { {1} over {2} } αt rSup { size 8{2} } } {} #" "=0+ left (0 "." "500" right ) left ("110"" rad/s" rSup { size 8{2} } right ) left (2 "." "00"" s" right ) rSup { size 8{2} } ="220"" rad" {} } } {}

Converting radians to revolutions gives

θ = 220 rad 1 rev 2π rad = 35.0 rev. size 12{θ= left ("220"" rad" right ) { {1" rev"} over {2π" rad"} } ="35" "." 0" rev."} {}

Solution for (d)

The number of meters of fishing line is x size 12{x} {} , which can be obtained through its relationship with θ size 12{θ} {} :

x = = 0.0450 m 220 rad = 9.90 m . size 12{x=rθ= left (0 "." "0450"" m" right ) left ("220"" rad" right )=9 "." "90"" m"} {}

Discussion

This example illustrates that relationships among rotational quantities are highly analogous to those among linear quantities. We also see in this example how linear and rotational quantities are connected. The answers to the questions are realistic. After unwinding for two seconds, the reel is found to spin at 220 rad/s, which is 2100 rpm. (No wonder reels sometimes make high-pitched sounds.) The amount of fishing line played out is 9.90 m, about right for when the big fish bites.

The figure shows a fishing reel, with radius equal to 4.5 centimeters. The direction of rotation of the reel is counterclockwise. The rotational quantities are theta, omega and alpha, and x, v, a are linear or translational quantities. The reel, fishing line, and the direction of motion have been separately indicated by curved arrows pointing toward those parts.
Fishing line coming off a rotating reel moves linearly. [link] and [link] consider relationships between rotational and linear quantities associated with a fishing reel.

Calculating the duration when the fishing reel slows down and stops

Now let us consider what happens if the fisherman applies a brake to the spinning reel, achieving an angular acceleration of 300 rad/s 2 size 12{"300"`"rad/s" rSup { size 8{2} } } {} . How long does it take the reel to come to a stop?

Strategy

We are asked to find the time t size 12{α} {} for the reel to come to a stop. The initial and final conditions are different from those in the previous problem, which involved the same fishing reel. Now we see that the initial angular velocity is ω 0 = 220 rad/s size 12{ω rSub { size 8{0} } ="220"" rad/s"} {} and the final angular velocity ω size 12{ω} {} is zero. The angular acceleration is given to be α = 300 rad/s 2 size 12{α= - "300" "rad/s" rSup { size 8{2} } } {} . Examining the available equations, we see all quantities but t are known in ω = ω 0 + αt , size 12{ω=ω rSub { size 8{0} } +αt} {} making it easiest to use this equation.

Solution

The equation states

ω = ω 0 + αt . size 12{ω=ω rSub { size 8{0} } +αt"."} {}

We solve the equation algebraically for t , and then substitute the known values as usual, yielding

t = ω ω 0 α = 0 220 rad/s 300 rad/s 2 = 0 . 733 s. size 12{t= { {ω - ω rSub { size 8{0} } } over {α} } = { {0 - "220"" rad/s"} over { - "300""rad/s" rSup { size 8{2} } } } =0 "." "733"" s."} {}

Discussion

Note that care must be taken with the signs that indicate the directions of various quantities. Also, note that the time to stop the reel is fairly small because the acceleration is rather large. Fishing lines sometimes snap because of the accelerations involved, and fishermen often let the fish swim for a while before applying brakes on the reel. A tired fish will be slower, requiring a smaller acceleration.

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Questions & Answers

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Source:  OpenStax, College physics for ap® courses. OpenStax CNX. Nov 04, 2016 Download for free at https://legacy.cnx.org/content/col11844/1.14
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