VIPSolutionsAct like a helpful tutor and exlain me : Give step-by-step solution with explanation and final answer:explain completeley , The engine and propeller of a small aeroplane rotates anticlockwise when viewed from the front. The combined moment of inertia of the engine and propeller system is 1 8 0 kg · m ^ 2 . The system operates steadily at 2 0 0 0 rpm . Calculate the magnitude and explain the effect of the gyroscopic couple produced when the aeroplane makes a right - hand turn with a turning radius of 3 0 0 0 meters at a forward speed of 5 0 0 km / h .
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Act like a helpful tutor and exlain me : Give step-by-step solution with explanation and final answer:explain completeley , The engine and propeller of a small aeroplane rotates anticlockwise when viewed from the front. The combined moment of inertia of the engine and propeller system is 1 8 0 kg · m ^ 2 . The system operates steadily at 2 0 0 0 rpm . Calculate the magnitude and explain the effect of the gyroscopic couple produced when the aeroplane makes a right - hand turn with a turning radius of 3 0 0 0 meters at a forward speed of 5 0 0 km / h .
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Asked by: manish kumar
Created at: 2025-10-21 20:46:14
Asked by: manish kumar
Created at: 2025-10-21 20:46:14
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# Gyroscopic Effect in an Aeroplane
## Given Data:
- Moment of inertia of engine and propeller, \( I = 180\, \text{kg·m}^2 \)
- Rotation speed, \( N = 200\, \text{rpm} \)
- Turning radius, \( R = 300\, \text{m} \)
- Forward speed, \( v = 50\, \text{km/h} \)
---
## Step 1: Convert rotational speed to radians per second
\[
N = 200\, \text{rpm}
\]
Convert rpm to radians per second:
\[
\omega = \frac{2 \pi N}{60}
\]
\[
\omega = \frac{2 \pi \times 200}{60} \approx \frac{2 \pi \times 200}{60} \approx 209.44\, \text{rad/sec}
\]
---
## Step 2: Calculate the angular momentum \( \mathbf{L} \)
\[
L = I \omega
\]
\[
L = 180 \times 209.44 \approx 37,599.2\, \text{kg·m}^2/\text{s}
\]
---
## Step 3: Find the rate of turn (angular velocity of the turn)
The plane makes a turn with radius \( R \) at speed \( v \):
Convert \( v \) to m/sec:
\[
v = 50\, \text{km/h} = \frac{50 \times 100}{360} \approx 13.89\, \text{m/sec}
\]
Angular velocity of the turn, \( \Omega \):
\[
\Omega = \frac{v}{R} = \frac{13.89}{300} \approx .00463\, \text{rad/sec}
\]
---
## Step 4: Calculate the gyroscopic couple \( \mathbf{C} \)
The gyroscopic couple is given by:
\[
C = L \times \Omega
\]
Since the gyroscopic couple acts perpendicular to the angular momentum vector, its magnitude is:
\[
C = 37,599.2 \times .00463 \approx 174.2\, \text{Nm}
\]
---
## **Final Answer:**
### **Magnitude of gyroscopic couple:**
\[
\boxed{
C \approx 174\, \text{Nm}
}
\]
---
## **Explanation of the effect:**
- The gyroscopic couple tends to resist the change in the orientation of the engine and propeller system.
- When the airplane makes a right-hand turn, the gyroscopic couple acts **perpendicular** to the direction of rotation.
- This results in a **precessional effect**: the airplane experiences a yawing or pitching moment, which pilots need to compensate for.
- Specifically, in a right turn, the gyroscopic couple tends to **tilt the nose of the plane upward or downward** depending on the direction of rotation and turn.
**Summary:**
- The gyroscopic couple's magnitude is approximately **174 Nm**.
- It influences the aircraft's handling during turns, requiring pilot adjustments to maintain the desired attitude.
---
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