A slender uniform bar of mass 21.5 kg and length 2.3 m rotates about an axis 1.05 m from its mass centre. Compute the centroidal mass moment of inertia, the inertia about the offset axis, the radius of gyration, and the kinetic energy at 4.5 rad/s.

Start with the thinking

• The transfer term m d² is always positive: the centroidal axis gives the smallest possible inertia.
• The parallel-axis theorem only transfers between parallel axes, and one of them must be centroidal.

Step-by-step solution

1. Formula

   $$I_c = \dfrac{mL^2}{12}$$

2. Substituting — I_c = 21.5(2.3)²/12 = 9.4779 kg·m²

3. Formula

   $$I = I_c + m d^2$$

4. Transfer term — m d² = 21.5(1.05)² = 23.7038 kg·m²

5. Substituting — I = 9.4779 + 23.7038 = 33.1817 kg·m²

6. Formula

   $$k = \sqrt{I/m}$$

7. Substituting

   $$k = \sqrt(33.1817/21.5) = 1.242 m$$

8. Kinetic energy

   $$T = ½I\omega^{2} = 0.5(33.1817)(4.5)^{2} = 336.0 J$$

Why the other options are there

• I = -14.2258 kg·m² (transfer term subtracted)
• I = 37.9117 kg·m² (end-axis formula misapplied)

Reference: FE Reference Handbook — Dynamics → Parallel-Axis Theorem

A slender uniform bar of mass 27.0 kg and length 1.5 m rotates about an axis 0.70 m from its mass centre. Compute the centroidal mass moment of inertia, the inertia about the offset axis, the radius of gyration, and the kinetic energy at 4.5 rad/s.

Start with the thinking

• The transfer term m d² is always positive: the centroidal axis gives the smallest possible inertia.
• The parallel-axis theorem only transfers between parallel axes, and one of them must be centroidal.

Step-by-step solution

1. Formula

   $$I_c = \dfrac{mL^2}{12}$$

2. Substituting — I_c = 27.0(1.5)²/12 = 5.0625 kg·m²

3. Formula

   $$I = I_c + m d^2$$

4. Transfer term — m d² = 27.0(0.70)² = 13.2300 kg·m²

5. Substituting — I = 5.0625 + 13.2300 = 18.2925 kg·m²

6. Formula

   $$k = \sqrt{I/m}$$

7. Substituting

   $$k = \sqrt(18.2925/27.0) = 0.823 m$$

8. Kinetic energy

   $$T = ½I\omega^{2} = 0.5(18.2925)(4.5)^{2} = 185.2 J$$

Why the other options are there

• I = -8.1675 kg·m² (transfer term subtracted)
• I = 20.2500 kg·m² (end-axis formula misapplied)

Reference: FE Reference Handbook — Dynamics → Parallel-Axis Theorem

A slender uniform bar of mass 14.0 kg and length 2.7 m rotates about an axis 0.20 m from its mass centre. Compute the centroidal mass moment of inertia, the inertia about the offset axis, the radius of gyration, and the kinetic energy at 10.5 rad/s.

Start with the thinking

• The transfer term m d² is always positive: the centroidal axis gives the smallest possible inertia.
• The parallel-axis theorem only transfers between parallel axes, and one of them must be centroidal.

Step-by-step solution

1. Formula

   $$I_c = \dfrac{mL^2}{12}$$

2. Substituting — I_c = 14.0(2.7)²/12 = 8.5050 kg·m²

3. Formula

   $$I = I_c + m d^2$$

4. Transfer term — m d² = 14.0(0.20)² = 0.5600 kg·m²

5. Substituting — I = 8.5050 + 0.5600 = 9.0650 kg·m²

6. Formula

   $$k = \sqrt{I/m}$$

7. Substituting

   $$k = \sqrt(9.0650/14.0) = 0.805 m$$

8. Kinetic energy

   $$T = ½I\omega^{2} = 0.5(9.0650)(10.5)^{2} = 499.7 J$$

Why the other options are there

• I = 7.9450 kg·m² (transfer term subtracted)
• I = 34.0200 kg·m² (end-axis formula misapplied)

Reference: FE Reference Handbook — Dynamics → Parallel-Axis Theorem

A slender uniform bar of mass 40.0 kg and length 3.0 m rotates about an axis 0.40 m from its mass centre. Compute the centroidal mass moment of inertia, the inertia about the offset axis, the radius of gyration, and the kinetic energy at 9.0 rad/s.

Start with the thinking

• The transfer term m d² is always positive: the centroidal axis gives the smallest possible inertia.
• The parallel-axis theorem only transfers between parallel axes, and one of them must be centroidal.

Step-by-step solution

1. Formula

   $$I_c = \dfrac{mL^2}{12}$$

2. Substituting — I_c = 40.0(3.0)²/12 = 30.0000 kg·m²

3. Formula

   $$I = I_c + m d^2$$

4. Transfer term — m d² = 40.0(0.40)² = 6.4000 kg·m²

5. Substituting — I = 30.0000 + 6.4000 = 36.4000 kg·m²

6. Formula

   $$k = \sqrt{I/m}$$

7. Substituting

   $$k = \sqrt(36.4000/40.0) = 0.954 m$$

8. Kinetic energy

   $$T = ½I\omega^{2} = 0.5(36.4000)(9.0)^{2} = 1,474 J$$

Why the other options are there

• I = 23.6000 kg·m² (transfer term subtracted)
• I = 120.0 kg·m² (end-axis formula misapplied)

Reference: FE Reference Handbook — Dynamics → Parallel-Axis Theorem

A slender uniform bar of mass 4.5 kg and length 1.1 m rotates about an axis 1.10 m from its mass centre. Compute the centroidal mass moment of inertia, the inertia about the offset axis, the radius of gyration, and the kinetic energy at 14.0 rad/s.

Start with the thinking

• The transfer term m d² is always positive: the centroidal axis gives the smallest possible inertia.
• The parallel-axis theorem only transfers between parallel axes, and one of them must be centroidal.

Step-by-step solution

1. Formula

   $$I_c = \dfrac{mL^2}{12}$$

2. Substituting — I_c = 4.5(1.1)²/12 = 0.4538 kg·m²

3. Formula

   $$I = I_c + m d^2$$

4. Transfer term — m d² = 4.5(1.10)² = 5.4450 kg·m²

5. Substituting — I = 0.4538 + 5.4450 = 5.8988 kg·m²

6. Formula

   $$k = \sqrt{I/m}$$

7. Substituting

   $$k = \sqrt(5.8988/4.5) = 1.145 m$$

8. Kinetic energy

   $$T = ½I\omega^{2} = 0.5(5.8988)(14.0)^{2} = 578.1 J$$

Why the other options are there

• I = -4.9913 kg·m² (transfer term subtracted)
• I = 1.8150 kg·m² (end-axis formula misapplied)

Reference: FE Reference Handbook — Dynamics → Parallel-Axis Theorem

A slender uniform bar of mass 8.5 kg and length 3.0 m rotates about an axis 1.05 m from its mass centre. Compute the centroidal mass moment of inertia, the inertia about the offset axis, the radius of gyration, and the kinetic energy at 6.5 rad/s.

Start with the thinking

• The transfer term m d² is always positive: the centroidal axis gives the smallest possible inertia.
• The parallel-axis theorem only transfers between parallel axes, and one of them must be centroidal.

Step-by-step solution

1. Formula

   $$I_c = \dfrac{mL^2}{12}$$

2. Substituting — I_c = 8.5(3.0)²/12 = 6.3750 kg·m²

3. Formula

   $$I = I_c + m d^2$$

4. Transfer term — m d² = 8.5(1.05)² = 9.3713 kg·m²

5. Substituting — I = 6.3750 + 9.3713 = 15.7463 kg·m²

6. Formula

   $$k = \sqrt{I/m}$$

7. Substituting

   $$k = \sqrt(15.7463/8.5) = 1.361 m$$

8. Kinetic energy

   $$T = ½I\omega^{2} = 0.5(15.7463)(6.5)^{2} = 332.6 J$$

Why the other options are there

• I = -2.9963 kg·m² (transfer term subtracted)
• I = 25.5000 kg·m² (end-axis formula misapplied)

Reference: FE Reference Handbook — Dynamics → Parallel-Axis Theorem

A slender uniform bar of mass 32.5 kg and length 2.3 m rotates about an axis 0.30 m from its mass centre. Compute the centroidal mass moment of inertia, the inertia about the offset axis, the radius of gyration, and the kinetic energy at 4.0 rad/s.

Start with the thinking

• The transfer term m d² is always positive: the centroidal axis gives the smallest possible inertia.
• The parallel-axis theorem only transfers between parallel axes, and one of them must be centroidal.

Step-by-step solution

1. Formula

   $$I_c = \dfrac{mL^2}{12}$$

2. Substituting — I_c = 32.5(2.3)²/12 = 14.3271 kg·m²

3. Formula

   $$I = I_c + m d^2$$

4. Transfer term — m d² = 32.5(0.30)² = 2.9250 kg·m²

5. Substituting — I = 14.3271 + 2.9250 = 17.2521 kg·m²

6. Formula

   $$k = \sqrt{I/m}$$

7. Substituting

   $$k = \sqrt(17.2521/32.5) = 0.729 m$$

8. Kinetic energy

   $$T = ½I\omega^{2} = 0.5(17.2521)(4.0)^{2} = 138.0 J$$

Why the other options are there

• I = 11.4021 kg·m² (transfer term subtracted)
• I = 57.3083 kg·m² (end-axis formula misapplied)

Reference: FE Reference Handbook — Dynamics → Parallel-Axis Theorem

A slender uniform bar of mass 17.0 kg and length 2.4 m rotates about an axis 0.55 m from its mass centre. Compute the centroidal mass moment of inertia, the inertia about the offset axis, the radius of gyration, and the kinetic energy at 2.5 rad/s.

Start with the thinking

• The transfer term m d² is always positive: the centroidal axis gives the smallest possible inertia.
• The parallel-axis theorem only transfers between parallel axes, and one of them must be centroidal.

Step-by-step solution

1. Formula

   $$I_c = \dfrac{mL^2}{12}$$

2. Substituting — I_c = 17.0(2.4)²/12 = 8.1600 kg·m²

3. Formula

   $$I = I_c + m d^2$$

4. Transfer term — m d² = 17.0(0.55)² = 5.1425 kg·m²

5. Substituting — I = 8.1600 + 5.1425 = 13.3025 kg·m²

6. Formula

   $$k = \sqrt{I/m}$$

7. Substituting

   $$k = \sqrt(13.3025/17.0) = 0.885 m$$

8. Kinetic energy

   $$T = ½I\omega^{2} = 0.5(13.3025)(2.5)^{2} = 41.57 J$$

Why the other options are there

• I = 3.0175 kg·m² (transfer term subtracted)
• I = 32.6400 kg·m² (end-axis formula misapplied)

Reference: FE Reference Handbook — Dynamics → Parallel-Axis Theorem

A slender uniform bar of mass 32.0 kg and length 0.9 m rotates about an axis 1.05 m from its mass centre. Compute the centroidal mass moment of inertia, the inertia about the offset axis, the radius of gyration, and the kinetic energy at 5.5 rad/s.

Start with the thinking

• The transfer term m d² is always positive: the centroidal axis gives the smallest possible inertia.
• The parallel-axis theorem only transfers between parallel axes, and one of them must be centroidal.

Step-by-step solution

1. Formula

   $$I_c = \dfrac{mL^2}{12}$$

2. Substituting — I_c = 32.0(0.9)²/12 = 2.1600 kg·m²

3. Formula

   $$I = I_c + m d^2$$

4. Transfer term — m d² = 32.0(1.05)² = 35.2800 kg·m²

5. Substituting — I = 2.1600 + 35.2800 = 37.4400 kg·m²

6. Formula

   $$k = \sqrt{I/m}$$

7. Substituting

   $$k = \sqrt(37.4400/32.0) = 1.082 m$$

8. Kinetic energy

   $$T = ½I\omega^{2} = 0.5(37.4400)(5.5)^{2} = 566.3 J$$

Why the other options are there

• I = -33.1200 kg·m² (transfer term subtracted)
• I = 8.6400 kg·m² (end-axis formula misapplied)

Reference: FE Reference Handbook — Dynamics → Parallel-Axis Theorem

A slender uniform bar of mass 10.0 kg and length 2.6 m rotates about an axis 0.65 m from its mass centre. Compute the centroidal mass moment of inertia, the inertia about the offset axis, the radius of gyration, and the kinetic energy at 15.0 rad/s.

Start with the thinking

• The transfer term m d² is always positive: the centroidal axis gives the smallest possible inertia.
• The parallel-axis theorem only transfers between parallel axes, and one of them must be centroidal.

Step-by-step solution

1. Formula

   $$I_c = \dfrac{mL^2}{12}$$

2. Substituting — I_c = 10.0(2.6)²/12 = 5.6333 kg·m²

3. Formula

   $$I = I_c + m d^2$$

4. Transfer term — m d² = 10.0(0.65)² = 4.2250 kg·m²

5. Substituting — I = 5.6333 + 4.2250 = 9.8583 kg·m²

6. Formula

   $$k = \sqrt{I/m}$$

7. Substituting

   $$k = \sqrt(9.8583/10.0) = 0.993 m$$

8. Kinetic energy

   $$T = ½I\omega^{2} = 0.5(9.8583)(15.0)^{2} = 1,109 J$$

Why the other options are there

• I = 1.4083 kg·m² (transfer term subtracted)
• I = 22.5333 kg·m² (end-axis formula misapplied)

Reference: FE Reference Handbook — Dynamics → Parallel-Axis Theorem