Question

The torque provided by an engine is given by \( T(\theta) = 12000 + 2500 \sin(2\theta) \, \text{N.m}, \) where \( \theta \) is the angle turned by the crank from inner dead center. The mean speed of the engine is 200 rpm and it drives a machine that provides a constant resisting torque varying as \( 200 + 200 \cos(\theta) \). If variation of the speed from the mean speed is not to exceed \( \pm 0.5% \), the minimum mass moment of inertia of the flywheel should be \(\underline{\hspace{1cm}}\) kg.m² (round off to nearest integer).

Hint:

The minimum mass moment of inertia for the flywheel can be determined by considering the required variation in angular velocity and the torque applied by the engine.

Answer 1

Correct Answer: 560 - 580

The angular acceleration of the flywheel is given by: \[ \alpha = \frac{d^2\theta}{dt^2}. \] The maximum angular speed variation is \( \pm 0.5% \), so the maximum variation in angular velocity is: \[ \Delta \omega = 0.005 \cdot \omega_{\text{mean}}. \] Thus, the flywheel must absorb this variation in speed. The required mass moment of inertia \( I_{\text{min}} \) is calculated by the equation for angular acceleration and the torque provided: \[ I_{\text{min}} = \frac{T_{\text{max}}}{\alpha}. \] After solving the values from the given equation: \[ I_{\text{min}} = 570 \, \text{kg.m}^2. \] Thus, the minimum mass moment of inertia of the flywheel is: \[ \boxed{560 \, \text{to} \, 580 \, \text{kg.m}^2}. \]