Angular Momentum or Moment of Momentum

Dynamics · FE Reference Handbook section

Dynamics

4 formulas

10 exam-style examples

~53 min

All Dynamics lectures

Handbook notes for this section

Definitions and conditions exactly as the handbook states them.

The angular momentum or the moment of momentum about point 0 for a particle is defined as
Taking the time derivative of the above, the equation of motion may be written as
where M0 is the moment applied to the particle. Now by integrating and summing over a system of any number of particles, this
The term on the left side of the equation is the angular momentum of a system of particles at time t2. The first term on the
right side of the equation is the angular momentum of a system of particles at time t1. The second term on the right side of the
equation is the angular impulse of the moment M0 from time t1 to t2.

Core formulas for this FE topic

Definitions, applicability, units, assumptions and worked examples for each relation.

Worked exam-style examples

The four ways this section is written on the real exam — thoughts first, then equations, then substitution.

Example 1

Perfectly plastic impact — Angular Momentum or Moment of Momentum

A 8 kg mass moving at 13 m/s strikes a stationary 3 kg mass and they move together. Find the common velocity and the energy lost.

Given

$m_{1} = 8 kg, v_{1} = 13 m/s$
$m_{2} = 3 kg, v_{2} = 0$

Find

v' and ΔKE

Start with the thinking

Momentum is conserved in every impact; energy is not.
Plastic impact means one common final velocity.

Step-by-step solution

Momentum
$m_{1}v_{1} + m_{2}v_{2} = (m_{1} + m_{2})v'$
Substituting
$8(13) + 3(0) = (11)v'$
Solve
$v' = 104/11 = 9.455 m/s$
Initial KE
$½(8)(13)^{2} = 676.0 J$
Final KE
$½(11)(9.455)^{2} = 491.6 J$
Energy lost — ΔKE = 184.4 J

Answer:

v′ ≈ 9.45 m/s; ΔKE ≈ 184.4 J

Why the other options are there

6.50 m/s (masses assumed equal)
ΔKE = 0 (energy assumed conserved)

Reference: FE Reference Handbook — Dynamics → Angular Momentum or Moment of Momentum

Example 2

Angular acceleration of a rotating drum — Angular Momentum or Moment of Momentum

A drum with mass moment of inertia 4.5 kg·m² is driven by a constant torque of 92 N·m from rest. Find α and the angular speed after 9.5 s.

Given

I = 4.5 kg·m²
M = 92 N·m
$t = 9.5 s$

Find

α and ω

Start with the thinking

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

Step-by-step solution

Rotational equation
$M = I\alpha$
Substituting
$\alpha = 92/4.5 = 20.444 rad/s^{2}$
Angular speed — ω = ω₀ + αt
Substituting
$\omega = 0 + 20.444(9.5) = 194.2 rad/s$
Convert
$\omega = 1,855 rpm$

Answer:

α ≈ 20.44 rad/s²; ω ≈ 194.2 rad/s (1,855 rpm)

Why the other options are there

414.0 rad/s² (multiplied instead of divided)
30.91 rad/s (revolutions confused with radians)

Reference: FE Reference Handbook — Dynamics → Angular Momentum or Moment of Momentum

Example 3

Perfectly plastic impact — Angular Momentum or Moment of Momentum (2)

A 3 kg mass moving at 25 m/s strikes a stationary 3 kg mass and they move together. Find the common velocity and the energy lost.

Given

$m_{1} = 3 kg, v_{1} = 25 m/s$
$m_{2} = 3 kg, v_{2} = 0$

Find

v' and ΔKE

Start with the thinking

Momentum is conserved in every impact; energy is not.
Plastic impact means one common final velocity.

Step-by-step solution

Momentum
$m_{1}v_{1} + m_{2}v_{2} = (m_{1} + m_{2})v'$
Substituting
$3(25) + 3(0) = (6)v'$
Solve
$v' = 75/6 = 12.500 m/s$
Initial KE
$½(3)(25)^{2} = 937.5 J$
Final KE
$½(6)(12.500)^{2} = 468.8 J$
Energy lost — ΔKE = 468.8 J

Answer:

v′ ≈ 12.50 m/s; ΔKE ≈ 468.8 J

Why the other options are there

12.50 m/s (masses assumed equal)
ΔKE = 0 (energy assumed conserved)

Reference: FE Reference Handbook — Dynamics → Angular Momentum or Moment of Momentum

Example 4

Angular acceleration of a rotating drum — Angular Momentum or Moment of Momentum (2)

A drum with mass moment of inertia 5.0 kg·m² is driven by a constant torque of 85 N·m from rest. Find α and the angular speed after 2.5 s.

Given

I = 5.0 kg·m²
M = 85 N·m
$t = 2.5 s$

Find

α and ω

Start with the thinking

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

Step-by-step solution

Rotational equation
$M = I\alpha$
Substituting
$\alpha = 85/5.0 = 17.000 rad/s^{2}$
Angular speed — ω = ω₀ + αt
Substituting
$\omega = 0 + 17.000(2.5) = 42.50 rad/s$
Convert
$\omega = 405.8 rpm$

Answer:

α ≈ 17.00 rad/s²; ω ≈ 42.5 rad/s (405.8 rpm)

Why the other options are there

425.0 rad/s² (multiplied instead of divided)
6.76 rad/s (revolutions confused with radians)

Reference: FE Reference Handbook — Dynamics → Angular Momentum or Moment of Momentum

Example 5

Perfectly plastic impact — Angular Momentum or Moment of Momentum (3)

A 6 kg mass moving at 25 m/s strikes a stationary 15 kg mass and they move together. Find the common velocity and the energy lost.

Given

$m_{1} = 6 kg, v_{1} = 25 m/s$
$m_{2} = 15 kg, v_{2} = 0$

Find

v' and ΔKE

Start with the thinking

Momentum is conserved in every impact; energy is not.
Plastic impact means one common final velocity.

Step-by-step solution

Momentum
$m_{1}v_{1} + m_{2}v_{2} = (m_{1} + m_{2})v'$
Substituting
$6(25) + 15(0) = (21)v'$
Solve
$v' = 150/21 = 7.143 m/s$
Initial KE
$½(6)(25)^{2} = 1,875 J$
Final KE
$½(21)(7.143)^{2} = 535.7 J$
Energy lost — ΔKE = 1,339 J

Answer:

v′ ≈ 7.14 m/s; ΔKE ≈ 1,339 J

Why the other options are there

12.50 m/s (masses assumed equal)
ΔKE = 0 (energy assumed conserved)

Reference: FE Reference Handbook — Dynamics → Angular Momentum or Moment of Momentum

Example 6

Angular acceleration of a rotating drum — Angular Momentum or Moment of Momentum (3)

A drum with mass moment of inertia 18.5 kg·m² is driven by a constant torque of 137 N·m from rest. Find α and the angular speed after 9.5 s.

Given

I = 18.5 kg·m²
M = 137 N·m
$t = 9.5 s$

Find

α and ω

Start with the thinking

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

Step-by-step solution

Rotational equation
$M = I\alpha$
Substituting
$\alpha = 137/18.5 = 7.405 rad/s^{2}$
Angular speed — ω = ω₀ + αt
Substituting
$\omega = 0 + 7.405(9.5) = 70.35 rad/s$
Convert
$\omega = 671.8 rpm$

Answer:

α ≈ 7.41 rad/s²; ω ≈ 70.4 rad/s (671.8 rpm)

Why the other options are there

2,535 rad/s² (multiplied instead of divided)
11.20 rad/s (revolutions confused with radians)

Reference: FE Reference Handbook — Dynamics → Angular Momentum or Moment of Momentum

Example 7

Perfectly plastic impact — Angular Momentum or Moment of Momentum (4)

A 7 kg mass moving at 17 m/s strikes a stationary 11 kg mass and they move together. Find the common velocity and the energy lost.

Given

$m_{1} = 7 kg, v_{1} = 17 m/s$
$m_{2} = 11 kg, v_{2} = 0$

Find

v' and ΔKE

Start with the thinking

Momentum is conserved in every impact; energy is not.
Plastic impact means one common final velocity.

Step-by-step solution

Momentum
$m_{1}v_{1} + m_{2}v_{2} = (m_{1} + m_{2})v'$
Substituting
$7(17) + 11(0) = (18)v'$
Solve
$v' = 119/18 = 6.611 m/s$
Initial KE
$½(7)(17)^{2} = 1,012 J$
Final KE
$½(18)(6.611)^{2} = 393.4 J$
Energy lost — ΔKE = 618.1 J

Answer:

v′ ≈ 6.61 m/s; ΔKE ≈ 618.1 J

Why the other options are there

8.50 m/s (masses assumed equal)
ΔKE = 0 (energy assumed conserved)

Reference: FE Reference Handbook — Dynamics → Angular Momentum or Moment of Momentum

Example 8

Angular acceleration of a rotating drum — Angular Momentum or Moment of Momentum (4)

A drum with mass moment of inertia 10.0 kg·m² is driven by a constant torque of 160 N·m from rest. Find α and the angular speed after 8.0 s.

Given

I = 10.0 kg·m²
M = 160 N·m
$t = 8.0 s$

Find

α and ω

Start with the thinking

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

Step-by-step solution

Rotational equation
$M = I\alpha$
Substituting
$\alpha = 160/10.0 = 16.000 rad/s^{2}$
Angular speed — ω = ω₀ + αt
Substituting
$\omega = 0 + 16.000(8.0) = 128.0 rad/s$
Convert
$\omega = 1,222 rpm$

Answer:

α ≈ 16.00 rad/s²; ω ≈ 128.0 rad/s (1,222 rpm)

Why the other options are there

1,600 rad/s² (multiplied instead of divided)
20.37 rad/s (revolutions confused with radians)

Reference: FE Reference Handbook — Dynamics → Angular Momentum or Moment of Momentum

Example 9

Perfectly plastic impact — Angular Momentum or Moment of Momentum (5)

A 7 kg mass moving at 12 m/s strikes a stationary 14 kg mass and they move together. Find the common velocity and the energy lost.

Given

$m_{1} = 7 kg, v_{1} = 12 m/s$
$m_{2} = 14 kg, v_{2} = 0$

Find

v' and ΔKE

Start with the thinking

Momentum is conserved in every impact; energy is not.
Plastic impact means one common final velocity.

Step-by-step solution

Momentum
$m_{1}v_{1} + m_{2}v_{2} = (m_{1} + m_{2})v'$
Substituting
$7(12) + 14(0) = (21)v'$
Solve
$v' = 84/21 = 4.000 m/s$
Initial KE
$½(7)(12)^{2} = 504.0 J$
Final KE
$½(21)(4.000)^{2} = 168.0 J$
Energy lost — ΔKE = 336.0 J

Answer:

v′ ≈ 4.00 m/s; ΔKE ≈ 336.0 J

Why the other options are there

6.00 m/s (masses assumed equal)
ΔKE = 0 (energy assumed conserved)

Reference: FE Reference Handbook — Dynamics → Angular Momentum or Moment of Momentum

Example 10

Angular acceleration of a rotating drum — Angular Momentum or Moment of Momentum (5)

A drum with mass moment of inertia 29.0 kg·m² is driven by a constant torque of 34 N·m from rest. Find α and the angular speed after 4.5 s.

Given

I = 29.0 kg·m²
M = 34 N·m
$t = 4.5 s$

Find

α and ω

Start with the thinking

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

Step-by-step solution

Rotational equation
$M = I\alpha$
Substituting
$\alpha = 34/29.0 = 1.172 rad/s^{2}$
Angular speed — ω = ω₀ + αt
Substituting
$\omega = 0 + 1.172(4.5) = 5.28 rad/s$
Convert
$\omega = 50.4 rpm$

Answer:

α ≈ 1.17 rad/s²; ω ≈ 5.3 rad/s (50 rpm)

Why the other options are there

986.0 rad/s² (multiplied instead of divided)
0.84 rad/s (revolutions confused with radians)

Reference: FE Reference Handbook — Dynamics → Angular Momentum or Moment of Momentum