(1)A rescue helicopter of mass 810 kg , supports itself in a stationary position by imparting a downward velocity, \( v \), to the air in a circle of radius 4.0 m . The density of the air is 1.20 kg \( \mathrm{~m}^{-3} \).

To analyze the situation of the rescue helicopter hovering in a stationary position, we can use the principles of fluid dynamics and the concept of thrust. ### Given Data: - Mass of the helicopter, \( m = 810 \, \text{kg} \) - Radius of the downward air velocity circle, \( r = 4.0 \, \text{m} \) - Density of air, \( \rho = 1.20 \, \text{kg/m}^3 \) ### Step 1: Calculate the Weight of the Helicopter The weight \( W \) of the helicopter can be calculated using the formula: \[ W = m \cdot g \] where \( g \) is the acceleration due to gravity (approximately \( 9.81 \, \text{m/s}^2 \)). \[ W = 810 \, \text{kg} \cdot 9.81 \, \text{m/s}^2 = 7965.3 \, \text{N} \] ### Step 2: Calculate the Thrust Required For the helicopter to hover, the thrust \( T \) generated by the rotor must equal the weight of the helicopter: \[ T = W = 7965.3 \, \text{N} \] ### Step 3: Calculate the Volume of Air Displaced The volume of air displaced by the rotor can be calculated using the area of the circular region through which the air is being pushed down. The area \( A \) of the circle is given by: \[ A = \pi r^2 \] Substituting the radius: \[ A = \pi (4.0 \, \text{m})^2 = \pi \cdot 16 \, \text{m}^2 \approx 50.27 \, \text{m}^2 \] ### Step 4: Calculate the Mass Flow Rate of Air The thrust generated by the helicopter can also be expressed in terms of the mass flow rate \( \dot{m} \) of the air and the velocity \( v \) of the air being pushed down: \[ T = \dot{m} \cdot v \] The mass flow rate \( \dot{m} \) can be expressed as: \[ \dot{m} = \rho \cdot A \cdot v \] Substituting this into the thrust equation gives: \[ T = \rho \cdot A \cdot v^2 \] ### Step 5: Solve for the Downward Velocity \( v \) Now we can rearrange the equation to solve for \( v \): \[ v^2 = \frac{T}{\rho \cdot A} \] Substituting the known values: \[ v^2 = \frac{7965.3 \, \text{N}}{1.20 \, \text{kg/m}^3 \cdot 50.27 \, \text{m}^2} \] Calculating the denominator: \[ 1.20 \cdot 50.27 \approx 60.324 \, \text{kg/m}^2 \] Now substituting back: \[ v^2 = \frac{7965.3}{60.324} \approx 132.4 \, \text{m}^2/\text{s}^2 \] Taking the square root to find \( v \): \[ v \approx \sqrt{132.4} \approx 11.5 \, \text{m/s} \] ### Conclusion The downward velocity \( v \) of the air imparted by the helicopter is approximately \( 11.5 \, \text{m/s} \).