A fish takes the bait and pulls on the line with a force of 2.2 N. The fishing reel, which has a friction clutch that exerts a restraining torque of 0.048 N m, is a cylinder of radius 0.055 m and mass 0.84 kg. Part A

What is the angular acceleration of the fishing reel?
B
How much line does the fish pull from the reel in 0.25 s?

To find the angular acceleration of the fishing reel, we can make use of the torque equation:

Torque = Moment of Inertia * Angular Acceleration

In this case, the torque exerted by the fish pulling on the line is equal to the torque exerted by the friction clutch of the fishing reel.

Given:
Torque by the fish pulling on the line = 2.2 N * 0.055 m (Force * Radius) = 0.121 N m
Torque exerted by the friction clutch = 0.048 N m
Moment of Inertia (I) of the fishing reel is given by the formula: I = (1/2) * Mass * Radius^2
Mass of the fishing reel = 0.84 kg
Radius of the fishing reel = 0.055 m

Substituting the values into the torque equation, we have:

0.121 N m - 0.048 N m = (1/2) * 0.84 kg * 0.055 m^2 * Angular Acceleration

0.073 N m = 0.0243 kg m^2 * Angular Acceleration

Angular Acceleration = 0.073 N m / 0.0243 kg m^2
Angular Acceleration ≈ 3 rad/s^2

Therefore, the angular acceleration of the fishing reel is approximately 3 rad/s^2.

To find the amount of line the fish pulls from the reel in 0.25 s, we can use the formula:

Theta (angle) = Initial Angular Velocity * Time + (1/2) * Angular Acceleration * Time^2

Since the initial angular velocity is zero, the equation simplifies to:

Theta = (1/2) * Angular Acceleration * Time^2

Given the time as 0.25 s and the angular acceleration as 3 rad/s^2, we can calculate:

Theta = (1/2) * 3 rad/s^2 * (0.25 s)^2
Theta = 0.094 rad

Therefore, the fish pulls approximately 0.094 radians of line from the reel in 0.25 seconds.