The torque provided by an engine is given by T(θ) = 12000 + 2500 sin (2θ) N.m, where θ is the angle turned by the crank from the inner dead centre. The mean speed of the engine is 200 rpm and it drives a machine that provides constant resisting torque. If variation of the speed from the mean speed is not to exceed ±0.5%, the minimum mass moment of inertia of the flywheel should be ________ kg.m² (round off to the nearest integer).

The correct answer is 569.98.

Based on the provided image, we can analyze the problem step-by-step to find the minimum mass moment of inertia of the flywheel. Let us extract the details given in the image:
1. Engine torque expression:
$$T(\theta )=12000+2500sin(2\theta )\text{ N·m}$$
2. Mean speed of the engine:
$$N=200\text{ rpm}$$
3. The speed variation from the mean speed is limited to ±0.5%.

Step 1: Calculate the mean angular speed ($\omega$)
Using the relation between rotational speed in rpm and angular velocity in rad/s:
$$\omega =\frac{2\pi N}{60}$$
Substituting $N=200$ rpm:
$$\omega =\frac{2\times \pi \times 200}{60}\approx 20.944\text{ rad/s}$$

Step 2: Find the coefficient of fluctuation of speed ($K_s$)
The speed variation is given as ±0.5%. The total fluctuation of speed from minimum to maximum is:
$$K_s=0.5\%-(-0.5\%)=1\%=0.01$$

Step 3: Calculate the mean resisting torque ($T_{\mathrm{av}}$)
Since the torque function is periodic with a period of π, the cycle angle is π. The mean resisting torque matches the average torque over one cycle:
$$T_{\mathrm{av}}=\frac{1}{\pi }{\int }_0^{\pi }(12000+2500sin2\theta )d\theta$$
Integrating the term, we get:
$$T_{\mathrm{av}}=\frac{1}{\pi }{[12000\theta -\frac{2500}{2}}_0^{\pi }=12000\text{ N·m}$$

Step 4: Find the crank angles where the engine torque equals the resisting torque
Set the engine torque equal to the average torque:
$$12000+2500sin2\theta =12000$$
This simplifies to:
$$2500sin2\theta =0\Rightarrow 2\theta =0,\pi ,2\pi$$
Thus, the transition angles over a single cycle are:
$${\theta }_a=0,{\theta }_b=\frac{\pi }{2},{\theta }_c=\pi$$

Step 5: Determine the maximum fluctuation of energy ($E_f$)
Let the energy at ${\theta }_a=0$ be $E_a=E^{\prime }$.
The energy at ${\theta }_b=\frac{\pi }{2}$ is:
$$E_b=E^{\prime }+{\int }_0^{\pi /2}(T-T_{\mathrm{av}})d\theta$$
$$E_b=E^{\prime }+{\int }_0^{\pi /2}2500sin(2\theta )d\theta$$
Evaluating this integral yields:
$$E_b=E^{\prime }+{[-1250cos2\theta ]}_0^{\pi /2}=E^{\prime }+(1250-(-1250))=E^{\prime }+2500$$
This is the maximum energy level. The minimum energy level is $E^{\prime }$ (occurring at $\theta =0$ and $\theta =\pi$).
Thus, the maximum fluctuation of energy is:
$$E_f=E_{\max }-E_{\min }=(E^{\prime }+2500)-E^{\prime }=2500\text{ N·m}$$

Step 6: Compute the moment of inertia ($I$)
The relation between maximum fluctuation of energy, moment of inertia, mean angular velocity, and coefficient of fluctuation of speed is:
$$E_f=I{\omega }^2{K}_s$$
Substituting the calculated values into the formula:
$$2500=I\times {(20.944)}^2\times 0.01$$
Solving for $I$:
$$I=\frac{2500}{{(20.944)}^2\times 0.01}$$
$$I=\frac{2500}{438.65\times 0.01}=\frac{2500}{4.3865}\approx 569.98{\text{ kg·m}}^2$$