An airplane is flying in a straight line with a velocity of 200 mi /h and an acceleration of 3 mi/h2 . If the propeller has a diameter of 6 ft and is rotating at an angular rate of 120 rad/s, determine the magnitudes of velocity and acceleration of a particle located on the tip of the propeller.
The correct answer and explanation is :
Problem Statement:
An airplane is flying in a straight line with a velocity of 200 mi/h and an acceleration of 3 mi/h². The propeller has a diameter of 6 ft and is rotating at an angular rate of 120 rad/s. We need to determine the magnitudes of velocity and acceleration of a particle located on the tip of the propeller.
Solution:
We will solve for the velocity and acceleration of a particle at the tip of the propeller by considering the two components: the velocity and acceleration due to the airplane’s motion and the velocity and acceleration due to the rotation of the propeller.
1. Tangential Velocity of the Propeller Tip:
The propeller is rotating at an angular velocity (( \omega )) of 120 rad/s. The diameter of the propeller is given as 6 feet, so the radius ( r ) is:
[
r = \frac{\text{Diameter}}{2} = \frac{6 \text{ ft}}{2} = 3 \text{ ft}
]
The tangential velocity ( v_{\text{tangential}} ) of the particle at the tip of the propeller is given by:
[
v_{\text{tangential}} = r \cdot \omega
]
Substituting the values:
[
v_{\text{tangential}} = 3 \text{ ft} \times 120 \text{ rad/s} = 360 \text{ ft/s}
]
To convert this to miles per hour (mi/h), we use the conversion factor ( 1 \text{ mi} = 5280 \text{ ft} ) and ( 1 \text{ hour} = 3600 \text{ s} ):
[
v_{\text{tangential}} = 360 \times \frac{3600}{5280} \approx 244.44 \text{ mi/h}
]
2. Tangential Acceleration of the Propeller Tip:
The tangential acceleration ( a_{\text{tangential}} ) is related to the angular acceleration ( \alpha ), which we can calculate from the given information. Assuming the propeller’s angular velocity is constant, there is no angular acceleration, so:
[
a_{\text{tangential}} = 0 \text{ mi/h}^2
]
3. Total Velocity of the Particle at the Tip of the Propeller:
The airplane is flying in a straight line with a velocity of 200 mi/h. The total velocity ( v_{\text{total}} ) at the tip of the propeller is the vector sum of the airplane’s velocity and the tangential velocity due to the rotation of the propeller. Since these two velocities are perpendicular to each other, we can use the Pythagorean theorem:
[
v_{\text{total}} = \sqrt{(v_{\text{airplane}})^2 + (v_{\text{tangential}})^2}
]
Substituting the values:
[
v_{\text{total}} = \sqrt{(200)^2 + (244.44)^2} \approx \sqrt{40000 + 59746.6} \approx \sqrt{99746.6} \approx 316.02 \text{ mi/h}
]
4. Total Acceleration of the Particle at the Tip of the Propeller:
The total acceleration ( a_{\text{total}} ) of the particle at the tip of the propeller is the vector sum of two components: the acceleration due to the airplane’s motion and the centripetal acceleration due to the propeller’s rotation.
The centripetal acceleration ( a_{\text{centripetal}} ) is given by:
[
a_{\text{centripetal}} = \frac{v_{\text{tangential}}^2}{r}
]
Substituting the values:
[
a_{\text{centripetal}} = \frac{(360)^2}{3} = \frac{129600}{3} = 43200 \text{ ft/s}^2
]
Now, convert this to mi/h²:
[
a_{\text{centripetal}} = 43200 \times \left(\frac{3600}{5280}\right)^2 \approx 45.38 \text{ mi/h}^2
]
The acceleration due to the airplane’s motion is given as 3 mi/h². Since the centripetal acceleration and the airplane’s acceleration are perpendicular, the total acceleration ( a_{\text{total}} ) is:
[
a_{\text{total}} = \sqrt{(3)^2 + (45.38)^2} \approx \sqrt{9 + 2069.6} \approx \sqrt{2078.6} \approx 45.57 \text{ mi/h}^2
]
Final Answer:
- The magnitude of the velocity of the particle located on the tip of the propeller is approximately 316.02 mi/h.
- The magnitude of the acceleration of the particle located on the tip of the propeller is approximately 45.57 mi/h².