Let us treat a helicopter rotor blade as a long, thin rod, as shown in Fig. 8–49. (a) If each of the three rotor helicopter blades is 3.75 m long and has a mass of 135 kg, calculate the moment of inertia of the three rotor blades about the axis of rotation. (b) How much torque must the motor apply to bring the blades from rest to a speed of 6.0 rev/s in 8.0 s?

FIGURE 8-49

Problem 43

The torque is the product of the moment of inertia and the square of the distance. For this problem, first, find the total moment of inertia and then find the angular acceleration to find the torque.

The initial angular velocity of the blades is \({\omega _1} = 0\).

The final angular velocity is \({\omega _2} = 6.0\;{\rm{rev/s}} = 6.0 \times 2\pi \;{\rm{rad/s}}\).

The length of the blades is \(L = 3.75\;{\rm{m}}\).

The time is taken by the helicopter to reach the final angular velocity is \(t = 8.0\;{\rm{s}}\).

The mass of each blade is \(m = 135\;{\rm{kg}}\).

You can assume the blades as uniform rods rotating about one end.

Let I be the moment of inertia of the blades, and \(\tau \) is the required torque.

The moment of inertia of one blade is \(\frac{1}{3}m{L^2}\).

The total moment of inertia of the three blades is

\(\begin{align}I &= 3 \times \frac{1}{3}m{L^2}\\ &= 3 \times \frac{1}{3} \times \left( {135\;{\rm{kg}}} \right) \times {\left( {3.75\;{\rm{m}}} \right)^2}\\ &= 1898.44\;{\rm{kg}} \cdot {{\rm{m}}^{\rm{2}}}\\ \approx 1.90 \times {10^3}\;{\rm{kg}} \cdot {{\rm{m}}^{\rm{2}}}\end{align}\).

Hence, the moment of inertia is \(1.90 \times {10^3}\;{\rm{kg}} \cdot {{\rm{m}}^{\rm{2}}}\).

Now, the angular acceleration of the rotor is

\(\begin{align}\alpha &= \frac{{{\omega _2} - {\omega _1}}}{t}\\ &= \frac{{6.0 \times 2\pi \;{\rm{rad/s}} - 0}}{{8.0\;{\rm{s}}}}\\ &= 4.71\;{\rm{rad/}}{{\rm{s}}^{\rm{2}}}\end{align}\)

The required value of the torque is

\(\begin{align}\tau &= I\alpha \\ &= \left( {1898.44\;{\rm{kg}} \cdot {{\rm{m}}^{\rm{2}}}} \right) \times \left( {4.71\;{\rm{rad/}}{{\rm{s}}^{\rm{2}}}} \right)\\ &= 8941.65\;{\rm{m}} \cdot {\rm{N}}\\ &= 8900\;{\rm{m}} \cdot {\rm{N}}\end{align}\).

Hence, the required torque is \(8900\;{\rm{m}} \cdot {\rm{N}}\).