Question

A fish takes the bait and pulls on the line with a force of 2.3 N. The fishing reel, which rotates without friction, is a uniform cylinder of radius 0.045 m and mass 0.82 kg. What is the angular acceleration of the fishing reel? How much line does the fish pull from the reel in 0.20 s?

Answer 1

Answer:

$\alpha=12.47\ rad.s^{-2}$

$l=0.011223\ m=11.223\ mm$

Explanation:

Given:

• radius of cylinder, $r=0.045\ m$

• mass of cylinder, $m=0.82\ kg$

• time observation, $t=0.2\ s$

• force of pull by the fish, $F=2.3\ N$

Angular acceleration is given as:

$\tau=I.\alpha$ ............(1)

Now as, $\tau=F.r$ ...................(2)

where:

$\tau=$ torque

$I=$moment of inertia

$\alpha=$ angular acceleration.

From eq (2)

$\tau=2.3\times 0.045$

$\tau=0.1035\ N.m$

For cylinder:

$I=\frac{1}{2} m.r^{2}$

$I=0.5\times 0.82\times 0.045^2$

$I=8.3025\times 10^{-4}\ kg.m^2$

Now using eq. (1):

$0.1035=8.3025\times 10^{-4}\times \alpha$

$\alpha=12.47\ rad.s^{-2}$

Revolution made in 0.2 seconds:

$\theta=\omega_i.t+\frac{1}{2} \alpha.t^2$

where:

$\omega_i=$ initial angular velocity = 0

$\theta=0+0.5\times 12.47\times 0.2^2$

$\theta=0.2494\ rad$

Hence the length of reel pulled out:

$l=\theta \times r$

$l=0.2494\times 0.045$

$l=0.011223\ m=11.223\ mm$

Answer 2

Explanation:

The given data is as follows.

     Pulling force on the reel is F = T = 2.3 N

     Mass of cylinder (m) = 0.82 kg

    Radius of cylinder (r) = 0.045 m

Formula for torque pulling force on the cylinder is as follows.

           $\tau = Fr$

                     = $I \times \alpha$

Moment of inertia of the cylinder (I) is as follows.

            I = $\frac{mr^{2}}{2}$

  $\alpha$ = angular acceleration of the cylinder

Hence,

                    Fr = $(\frac{mr^{2}}{2}) \alpha$

or,   $\alpha = \frac{2F}{mr}$

                   = $\frac{2 \times 2.3 N}{0.82 kg \times 0.045 m}$

                   = 124.66 $rad/s^{2}$

Amount of line pulled is h and it is in a times of 0.2 sec.

Now, linear acceleration is calculated as follows.

             a = $r \alpha$

                = $0.045 m \times 124.66 rad/s^{2}$

                 = 5.61 $m/s^{2}$

Now, relation between h and acceleration as follows.

             h = $v_{o}t + \frac{1}{2}at^{2}$

here,   $v_{o}$ = 0

Hence, calculate the value of h as follows.

             h = $v_{o}t + \frac{1}{2}at^{2}$

                = $0 + \frac{1}{2} \times 5.61 m/s^{2} \times (0.2s)^{2}$

                = 0.1121 m

Thus, we can conclude that acceleration of the fishing reel is 5.61 $m/s^{2}$ and it will pull 5.61 $m/s^{2}$ line from the reel in 0.20 s.