A chain pulls tangentially on a \(40.6-\mathrm{kg}\) uniform cylindrical gear with a tension of 72.5 N. The chain is attached along the outside radius of the gear at \(0.650 \mathrm{m}\) from the axis of rotation. Starting from rest, the gear takes 1.70 s to reach its rotational speed of 1.35 rev/s. What is the total frictional torque opposing the rotation of the gear?

The tension in the chain applies a force on the gear tangentially, which causes a torque. The net torque can be calculated using the formula: $$ \tau_{net} = F_t \times r $$ where \(F_t\) is the tension in the chain and \(r\) is the radius of the gear. We are given \(F_t = 72.5\,\text{N}\) and \(r = 0.65\,\text{m}\), so we can calculate the net torque: $$ \tau_{net} = (72.5\,\text{N}) \times (0.65\,\text{m}) $$

We need to find the angular acceleration before we can find the frictional torque. Using the initial and final angular velocities and the time taken, we can apply the following formula to find the angular acceleration: $$ \alpha = \frac{\omega_f - \omega_i}{t} $$ where \(\alpha\) is the angular acceleration, \(\omega_f\) is the final angular velocity (given in rev/s, which we need to convert to rad/s), \(\omega_i\) is the initial angular velocity (which is zero as the gear starts from rest), and \(t\) is the time taken. We are given \(\omega_f = 1.35\, \text{rev/s}\). To convert it to rad/s, we multiply it by \(2\pi\): $$ \omega_f = 1.35\, \text{rev/s} \times 2\pi\, \frac{\text{rad}}{\text{rev}} = 8.498\, \text{rad/s} $$ Now, we have \(\omega_f = 8.498\,\text{rad/s}\), \(\omega_i = 0\,\text{rad/s}\), and \(t = 1.7\, \text{s}\), we can find the angular acceleration: $$ \alpha = \frac{8.498\,\text{rad/s} - 0\,\text{rad/s}}{1.7\,\text{s}} $$

We need to calculate the moment of inertia of the gear to determine the total frictional torque. As the gear is a uniform solid cylinder, its moment of inertia is given by: $$ I = \frac{1}{2}mr^2 $$ where \(I\) is the moment of inertia, \(m\) is the mass of the gear, and \(r\) is its radius. We are given \(m = 40.6\,\text{kg}\) and \(r = 0.65\,\text{m}\), so we can calculate the moment of inertia: $$ I = \frac{1}{2}(40.6\,\text{kg})(0.65\,\text{m})^2 $$

Now, we can use the moment of inertia and the angular acceleration to calculate the total frictional torque on the gear. Using Newton's second law for rotational motion, we can write: $$ \tau_{net} = I\alpha + \tau_{friction} $$ Substituting the values we obtained earlier, and solving for the frictional torque, \(\tau_{friction}\), we have: $$ \tau_{friction} = \tau_{net} - I\alpha $$ Compute the values to find the total frictional torque acting on the gear.