Sure! Let's solve the problem step-by-step.
### Given Data:
- Weight density of the material (\(\( \rho \)\)): \(\( 25000 \, \text{N/m}^3 \)\)
- Length of the shaft (\(\( L \)\)): \(\( 150 \, \text{cm} = 1.5 \, \text{m} \)\)
- Final speed (\(\( \omega \)\)): \(\( 750 \, \text{RPM} \)\)
- Time (\(\( t \)\)): \(\( 45 \, \text{seconds} \)\)
We need to convert the final speed from RPM to radians per second:
\(\[ \text{Final speed} = \omega = 750 \times \frac{2 \pi}{60} \, \text{rad/s} = 78.54 \, \text{rad/s} \]\)
The angular acceleration \(\( \alpha \)\) can be calculated as:
\(\[ \alpha = \frac{\omega}{t} = \frac{78.54}{45} = 1.745 \, \text{rad/s}^2 \]\)
### (i) Solid Shaft of Diameter 20cm
1. Diameter and Radius:
- Diameter (\(\( D \)\)): \(\( 20 \, \text{cm} = 0.2 \, \text{m} \)\)
- Radius (\(\( r \)\)): \(\( \frac{D}{2} = 0.1 \, \text{m} \)\)
2. Volume of the Solid Shaft:
- Volume (\(\( V \)\)) of a cylinder:
\(\[ V_{\text{solid}} = \pi r^2 L = \pi \times (0.1)^2 \times 1.5 = 0.0471 \, \text{m}^3 \]\)
3. Mass of the Solid Shaft:
- Weight density (\(\( w \)\)):
\(\[ \text{Weight} = \rho \times V \]\)
- Converting weight to mass (\(\( m \)\)):
\(\[ m_{\text{solid}} = \frac{\text{Weight}}{g} = \frac{25000 \times 0.0471}{9.81} = 120.09 \, \text{kg} \]\)
4. Moment of Inertia:
- For a solid cylinder:
\(\[ I_{\text{solid}} = 0.5 \times m \times r^2 = 0.5 \times 120.09 \times (0.1)^2 = 0.600 \, \text{kg} \cdot \text{m}^2 \]\)
5. Torque:
\(\[ \tau_{\text{solid}} = I \times \alpha = 0.600 \times 1.745 = 1.048 \, \text{N} \cdot \text{m} \]\)
### (ii) Hollow Shaft with Inner Diameter of 15cm and Outer Diameter of 20cm
1. Diameters and Radii:
- Outer diameter (\(\( D_{\text{outer}} \)\)): \(\( 20 \, \text{cm} = 0.2 \, \text{m} \)\)
- Inner diameter (\(\( D_{\text{inner}} \)\)): \(\( 15 \, \text{cm} = 0.15 \, \text{m} \)\)
- Outer radius (\(\( R \)\)): \(\( \frac{D_{\text{outer}}}{2} = 0.1 \, \text{m} \)\)
- Inner radius (\(\( r \)\)): \(\( \frac{D_{\text{inner}}}{2} = 0.075 \, \text{m} \)\)
2. Volume of the Hollow Shaft:
- Volume of a hollow cylinder:
\(\[ V_{\text{hollow}} = \pi L (R^2 - r^2) = \pi \times 1.5 \times ((0.1)^2 - (0.075)^2) = 0.0206 \, \text{m}^3 \]\)
3. Mass of the Hollow Shaft:
- Weight density (\(\( w \)\)):
\(\[ \text{Weight} = \rho \times V \]\)
- Converting weight to mass (\(\( m \)\)):
\(\[ m_{\text{hollow}} = \frac{\text{Weight}}{g} = \frac{25000 \times 0.0206}{9.81} = 52.54 \, \text{kg} \]\)
4. Moment of Inertia:
- For a hollow cylinder:
\(\[ I_{\text{hollow}} = 0.5 \times m \times (R^2 + r^2) = 0.5 \times 52.54 \times ((0.1)^2 + (0.075)^2) = 0.410 \, \text{kg} \cdot \text{m}^2 \]\)
5. Torque:
\(\[ \tau_{\text{hollow}} = I \times \alpha = 0.410 \times 1.745 = 0.716 \, \text{N} \cdot \text{m} \]\)
### Summary
The force (torque) required to accelerate the shaft:
- Solid Shaft:
- Volume: \(\(0.0471 \, \text{m}^3 \)\)
- Mass: \(\( 120.09 \, \text{kg} \)\)
- Moment of inertia: \(\( 0.600 \, \text{kg} \cdot \text{m}^2 \)\)
- Torque: \(\( 1.048 \, \text{N} \cdot \text{m} \)\)
- Hollow Shaft:
- Volume: \(\( 0.0206 \, \text{m}^3 \)\)
- Mass: \(\( 52.54 \, \text{kg} \)\)
- Moment of inertia: \(\( 0.410 \, \text{kg} \cdot \text{m}^2 \)\)
- Torque: \(\( 0.716 \, \text{N} \cdot \text{m} \)\)
