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

`\alpha=12.47\ rad.s^(-2)`

`l=0.011223\ m=11.223\ mm`

Step-by-step 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* 0.045`

`\tau=0.1035\ N.m`

For cylinder:

`I=(1)/(2) m.r^(2)`

`I=0.5* 0.82* 0.045^2`

`I=8.3025* 10^(-4)\ kg.m^2`

Now using eq. (1):

`0.1035=8.3025* 10^(-4)* \alpha`

`\alpha=12.47\ rad.s^(-2)`

Revolution made in 0.2 seconds:

`\theta=\omega_i.t+(1)/(2) \alpha.t^2`

where:

`\omega_i=` initial angular velocity = 0

`\theta=0+0.5* 12.47* 0.2^2`

`\theta=0.2494\ rad`

Hence the length of reel pulled out:

`l=\theta * r`

`l=0.2494* 0.045`

`l=0.011223\ m=11.223\ mm`

Answer 2

Step-by-step 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 * \alpha`

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

I =
`(mr^(2))/(2)`

`\alpha` = angular acceleration of the cylinder

Hence,

Fr =
`((mr^(2))/(2)) \alpha`

or,
`\alpha = (2F)/(mr)`

=
`(2 * 2.3 N)/(0.82 kg * 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 * 124.66 rad/s^(2)`

= 5.61
`m/s^(2)`

Now, relation between h and acceleration as follows.

h =
`v_(o)t + (1)/(2)at^(2)`

here,
`v_(o)` = 0

Hence, calculate the value of h as follows.

h =
`v_(o)t + (1)/(2)at^(2)`

=
`0 + (1)/(2) * 5.61 m/s^(2) * (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.