Question

The flywheel of an engine has moment of inertia 2.50 kg m2 about its rotation axis. What constant torque is required to bring it up to an angular speed of 400 rev/min in 8.00s, starting from rest?

Answer 1

Answer:

Explanation:

From the question we are told that

   The moment of inertia is  $I = 2.50 \ kg \cdot m^2$

    The final  angular speed is $w_f = 400 rev/min = \frac{400 * 2\pi}{60} = 41.89 \ rad/s$

     The time taken is  $t = 8.0 s$

      The initial angular speed is  $w_i = 0\ rad/s$

Generally the average angular acceleration is mathematically represented as

        $\alpha = \frac{w_f - w_i }{t}$

=>     $\alpha = \frac{41.89}{8}$

=>      $\alpha = 5.24 \ rad/s^2$

Generally the torque is mathematically represented as

   $\tau = I * \alpha$

=>    $\tau = 5.24 * 2.50$

=>     $\tau = 13.09 \ N \cdot m$