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