Question

a fish takes the bait and pulls on the line with a force of 2.2 n. the fishing reel, which rotates without friction, is a cylinder of radius 0.055 m and mass 0.99 kg. what is the angular acceleration of the fishing reel?

a. 7 rad/s²
b. 14 rad/s²
c. 23 rad/s²
d. 35 rad/s²
e. 81 rad/s²

Explanation:

Step1: Find the moment of inertia of the cylinder

The moment of inertia $I$ of a solid - cylinder is given by $I=\frac{1}{2}mr^{2}$, where $m = 0.99\ kg$ and $r=0.055\ m$.
$I=\frac{1}{2}\times0.99\times(0.055)^{2}=\frac{1}{2}\times0.99\times0.003025 = 0.001497375\ kg\cdot m^{2}$

Step2: Calculate the torque

The torque $\tau$ is given by $\tau = rF$, where $F = 2.2\ N$ and $r = 0.055\ m$.
$\tau=0.055\times2.2=0.121\ N\cdot m$

Step3: Use the relation between torque, moment of inertia and angular acceleration

The relation is $\tau = I\alpha$, where $\alpha$ is the angular acceleration. So, $\alpha=\frac{\tau}{I}$.
$\alpha=\frac{0.121}{0.001497375}\approx81\ rad/s^{2}$

Answer:

e. $81\ rad/s^{2}$