A 5858 lb cart is pulled horizontally by a 858 lb force against rolling resistance μ = 0.25. What is its acceleration?

Start with the thinking

• Convert weight to mass before applying F = ma.
• Net force is applied force minus resistance.

Step-by-step solution

1. Mass

   $$m = W/g = 5858/32.2 = 181.9 slugs$$

2. Resistance — F_r = μW = 0.25(5858) = 1,465 lb

3. Net force — ΣF = F − F_r

4. Substituting — ΣF = 858 − 1,465 = -606.5 lb

5. Acceleration — a = ΣF/m = -606.5/181.9 = -3.334 ft/s²

Why the other options are there

• -0.104 ft/s² (weight used as mass)
• 4.72 ft/s² (resistance ignored)

Reference: FE Reference Handbook — Dynamics → Mass Moment of Inertia

A drum with mass moment of inertia 10.0 kg·m² is driven by a constant torque of 69 N·m from rest. Find α and the angular speed after 2.0 s.

Start with the thinking

• Rotational analogue of F = ma is M = Iα.
• Constant torque gives constant angular acceleration.

Step-by-step solution

1. Rotational equation

   $$M = I\alpha$$

2. Substituting

   $$\alpha = 69/10.0 = 6.900 rad/s^{2}$$

3. Angular speed — ω = ω₀ + αt

4. Substituting

   $$\omega = 0 + 6.900(2.0) = 13.80 rad/s$$

5. Convert

   $$\omega = 131.8 rpm$$

Why the other options are there

• 690.0 rad/s² (multiplied instead of divided)
• 2.20 rad/s (revolutions confused with radians)

Reference: FE Reference Handbook — Dynamics → Mass Moment of Inertia

A slender uniform bar of mass 32.5 kg and length 2.9 m rotates about an axis 0.85 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 6.0 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(2.9)²/12 = 22.7771 kg·m²

3. Formula

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

4. Transfer term — m d² = 32.5(0.85)² = 23.4813 kg·m²

5. Substituting — I = 22.7771 + 23.4813 = 46.2583 kg·m²

6. Formula

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

7. Substituting

   $$k = \sqrt(46.2583/32.5) = 1.193 m$$

8. Kinetic energy

   $$T = ½I\omega^{2} = 0.5(46.2583)(6.0)^{2} = 832.7 J$$

Why the other options are there

• I = -0.7042 kg·m² (transfer term subtracted)
• I = 91.1083 kg·m² (end-axis formula misapplied)

Reference: FE Reference Handbook — Dynamics → Mass Moment of Inertia

A 2450 lb cart is pulled horizontally by a 679 lb force against rolling resistance μ = 0.25. What is its acceleration?

Start with the thinking

• Convert weight to mass before applying F = ma.
• Net force is applied force minus resistance.

Step-by-step solution

1. Mass

   $$m = W/g = 2450/32.2 = 76.09 slugs$$

2. Resistance — F_r = μW = 0.25(2450) = 612.5 lb

3. Net force — ΣF = F − F_r

4. Substituting — ΣF = 679 − 612.5 = 66.5 lb

5. Acceleration — a = ΣF/m = 66.5/76.09 = 0.874 ft/s²

Why the other options are there

• 0.027 ft/s² (weight used as mass)
• 8.92 ft/s² (resistance ignored)

Reference: FE Reference Handbook — Dynamics → Mass Moment of Inertia

A drum with mass moment of inertia 24.5 kg·m² is driven by a constant torque of 188 N·m from rest. Find α and the angular speed after 4.5 s.

Start with the thinking

• Rotational analogue of F = ma is M = Iα.
• Constant torque gives constant angular acceleration.

Step-by-step solution

1. Rotational equation

   $$M = I\alpha$$

2. Substituting

   $$\alpha = 188/24.5 = 7.673 rad/s^{2}$$

3. Angular speed — ω = ω₀ + αt

4. Substituting

   $$\omega = 0 + 7.673(4.5) = 34.53 rad/s$$

5. Convert

   $$\omega = 329.7 rpm$$

Why the other options are there

• 4,606 rad/s² (multiplied instead of divided)
• 5.50 rad/s (revolutions confused with radians)

Reference: FE Reference Handbook — Dynamics → Mass Moment of Inertia

A slender uniform bar of mass 37.5 kg and length 0.7 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 14.5 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 = 37.5(0.7)²/12 = 1.5313 kg·m²

3. Formula

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

4. Transfer term — m d² = 37.5(0.55)² = 11.3438 kg·m²

5. Substituting — I = 1.5313 + 11.3438 = 12.8750 kg·m²

6. Formula

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

7. Substituting

   $$k = \sqrt(12.8750/37.5) = 0.586 m$$

8. Kinetic energy

   $$T = ½I\omega^{2} = 0.5(12.8750)(14.5)^{2} = 1,353 J$$

Why the other options are there

• I = -9.8125 kg·m² (transfer term subtracted)
• I = 6.1250 kg·m² (end-axis formula misapplied)

Reference: FE Reference Handbook — Dynamics → Mass Moment of Inertia

A 4984 lb cart is pulled horizontally by a 332 lb force against rolling resistance μ = 0.20. What is its acceleration?

Start with the thinking

• Convert weight to mass before applying F = ma.
• Net force is applied force minus resistance.

Step-by-step solution

1. Mass

   $$m = W/g = 4984/32.2 = 154.8 slugs$$

2. Resistance — F_r = μW = 0.20(4984) = 996.8 lb

3. Net force — ΣF = F − F_r

4. Substituting — ΣF = 332 − 996.8 = -664.8 lb

5. Acceleration — a = ΣF/m = -664.8/154.8 = -4.295 ft/s²

Why the other options are there

• -0.133 ft/s² (weight used as mass)
• 2.14 ft/s² (resistance ignored)

Reference: FE Reference Handbook — Dynamics → Mass Moment of Inertia

A drum with mass moment of inertia 25.0 kg·m² is driven by a constant torque of 147 N·m from rest. Find α and the angular speed after 5.5 s.

Start with the thinking

• Rotational analogue of F = ma is M = Iα.
• Constant torque gives constant angular acceleration.

Step-by-step solution

1. Rotational equation

   $$M = I\alpha$$

2. Substituting

   $$\alpha = 147/25.0 = 5.880 rad/s^{2}$$

3. Angular speed — ω = ω₀ + αt

4. Substituting

   $$\omega = 0 + 5.880(5.5) = 32.34 rad/s$$

5. Convert

   $$\omega = 308.8 rpm$$

Why the other options are there

• 3,675 rad/s² (multiplied instead of divided)
• 5.15 rad/s (revolutions confused with radians)

Reference: FE Reference Handbook — Dynamics → Mass Moment of Inertia

A slender uniform bar of mass 23.0 kg and length 1.7 m rotates about an axis 1.15 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 = 23.0(1.7)²/12 = 5.5392 kg·m²

3. Formula

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

4. Transfer term — m d² = 23.0(1.15)² = 30.4175 kg·m²

5. Substituting — I = 5.5392 + 30.4175 = 35.9567 kg·m²

6. Formula

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

7. Substituting

   $$k = \sqrt(35.9567/23.0) = 1.250 m$$

8. Kinetic energy

   $$T = ½I\omega^{2} = 0.5(35.9567)(9.0)^{2} = 1,456 J$$

Why the other options are there

• I = -24.8783 kg·m² (transfer term subtracted)
• I = 22.1567 kg·m² (end-axis formula misapplied)

Reference: FE Reference Handbook — Dynamics → Mass Moment of Inertia

A 2732 lb cart is pulled horizontally by a 502 lb force against rolling resistance μ = 0.30. What is its acceleration?

Start with the thinking

• Convert weight to mass before applying F = ma.
• Net force is applied force minus resistance.

Step-by-step solution

1. Mass

   $$m = W/g = 2732/32.2 = 84.84 slugs$$

2. Resistance — F_r = μW = 0.30(2732) = 819.6 lb

3. Net force — ΣF = F − F_r

4. Substituting — ΣF = 502 − 819.6 = -317.6 lb

5. Acceleration — a = ΣF/m = -317.6/84.84 = -3.743 ft/s²

Why the other options are there

• -0.116 ft/s² (weight used as mass)
• 5.92 ft/s² (resistance ignored)

Reference: FE Reference Handbook — Dynamics → Mass Moment of Inertia