I have a disc 1200mm diameter, 16mm thickness, spinning at 850rpm. Weighing - 140Kg This disc is fixed to a 75mm diameter shaft. What radial force is generated by gyroscopic precession on the shaft if the shaft and disc is moved with a lever arm of 860mm in length at a speed of 0.5 m/s

To calculate the radial force generated by gyroscopic precession on the shaft, we need to apply the formula for gyroscopic torque and then convert that torque into a radial force. Here's how we approach the problem step by step:

### Step 1: Gyroscopic Torque Formula
The formula for the gyroscopic torque (\( T_g \)) is:

\[
T_g = I \cdot \omega \cdot \Omega
\]

Where:
- \( I \) = moment of inertia of the disc about its axis of rotation (kg·m²),
- \( \omega \) = angular velocity of the disc (rad/s),
- \( \Omega \) = angular velocity due to precession (rad/s).

### Step 2: Moment of Inertia (\( I \))
The disc is assumed to be a solid disc, so the moment of inertia (\( I \)) about its central axis is given by:

\[
I = \frac{1}{2} m r^2
\]

Where:
- \( m \) = mass of the disc = 140 kg,
- \( r \) = radius of the disc = \( \frac{1200 \, \text{mm}}{2} = 0.6 \, \text{m} \).

Substituting these values:

\[
I = \frac{1}{2} \times 140 \, \text{kg} \times (0.6 \, \text{m})^2 = 25.2 \, \text{kg·m}^2
\]

### Step 3: Angular Velocity of the Disc (\( \omega \))
The angular velocity (\( \omega \)) is related to the rotational speed by:

\[
\omega = \frac{2 \pi N}{60}
\]

Where:
- \( N \) = rotational speed in rpm = 850 rpm.

Substituting this:

\[
\omega = \frac{2 \pi \times 850}{60} \approx 89.01 \, \text{rad/s}
\]

### Step 4: Precession Angular Velocity (\( \Omega \))
The precession angular velocity (\( \Omega \)) is the angular velocity induced by the movement of the shaft. This can be calculated as:

\[
\Omega = \frac{v}{L}
\]

Where:
- \( v \) = speed of the lever arm = 0.5 m/s,
- \( L \) = length of the lever arm = 860 mm = 0.86 m.

Substituting these values:

\[
\Omega = \frac{0.5 \, \text{m/s}}{0.86 \, \text{m}} \approx 0.5814 \, \text{rad/s}
\]

### Step 5: Gyroscopic Torque (\( T_g \))

Now we can calculate the gyroscopic torque:

\[
T_g = 25.2 \, \text{kg·m}^2 \times 89.01 \, \text{rad/s} \times 0.5814 \, \text{rad/s} \approx 1303.93 \, \text{N·m}
\]

### Step 6: Radial Force

To find the radial force, we use the relationship:

\[
F = \frac{T_g}{r_{\text{shaft}}}
\]

Where:
- \( r_{\text{shaft}} \) = radius of the shaft = \( \frac{75 \, \text{mm}}{2} = 0.0375 \, \text{m} \).

Substituting the values:

\[
F = \frac{1303.93 \, \text{N·m}}{0.0375 \, \text{m}} \approx 34,771.47 \, \text{N}
\]

### Final Answer:
The radial force generated by gyroscopic precession on the shaft is approximately **34,771 N**.

Would you like more details or clarifications on any step?

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Here are 5 relative questions:
1. How does the diameter of the shaft affect the radial force?
2. How would changing the rotational speed (rpm) affect the gyroscopic force?
3. What happens if the mass of the disc is reduced to half?
4. How is the torque affected if the precession speed increases?
5. What effect would a thicker disc have on the gyroscopic precession?

**Tip**: The faster the disc spins, the greater the gyroscopic torque, but small changes in shaft size also drastically affect the radial force.

Learn how to calculate the radial force generated by gyroscopic precession on a disc with a diameter of 1200mm, spinning at 850rpm and weighing 140Kg. This detailed solution involves angular velocity, moment of inertia, and gyroscopic torque formulas. Suitable for advanced physics students.

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