Rigid Body Motion About a Fixed Axis

Example 1

Constant-acceleration kinematics — solve for final velocity — Rigid Body Motion About a Fixed Axis

A dynamics problem uses Constant-acceleration kinematics. Given initial velocity (v0) = 76.0000 ft/s; acceleration (a) = -12.0000 ft/s²; distance (s) = 259.0 ft, determine the final velocity (v) in ft/s.

Given

$initial velocity (v_{0}) = 76.0000 ft/s$
$acceleration (a) = -12.0000 ft/s^{2}$
$distance (s) = 259.0 ft$

Find

final velocity (v), in ft/s

Start with the thinking

The governing relation printed in this handbook section is Constant-acceleration kinematics.
Everything except v is given, so isolate v symbolically first — never rearrange after the numbers are in.
Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
Dynamics items reward recognising the unknown before touching a calculator.

Step-by-step solution

Step 1 — State the governing relation:
$v^2 = v_0^2 + 2 a s$
Step 2 — Rearrange the relation so that v stands alone on the left-hand side.
Step 3 — List the givens: initial velocity (v0) = 76.0000 ft/s, acceleration (a) = -12.0000 ft/s², distance (s) = 259.0 ft.
Step 4 — Substitute the given values into the rearranged relation.
Step 5 — Evaluate:
$v = 0.0000\ \text{ft/s}$
Step 6 — Check: returning v = 0.0000 ft/s to
$v^2 = v_0^2 + 2 a s$
reproduces the given quantities, and both sides carry the same units.

Answer:

v = 0.0000\ \text{ft/s}

Why the other options are there

0.0000 — kept a factor of two that cancels in the correct rearrangement.
0.0000 — dropped that same factor in the other direction.
0.0000 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Dynamics → Rigid Body Motion About a Fixed Axis

Example 2

Constant-acceleration kinematics — solve for acceleration — Rigid Body Motion About a Fixed Axis (2)

A dynamics problem uses Constant-acceleration kinematics. Given initial velocity (v0) = 36.0000 ft/s; distance (s) = 331.0 ft; final velocity (v) = 88.3000 ft/s, determine the acceleration (a) in ft/s².

Given

$initial velocity (v_{0}) = 36.0000 ft/s$
$distance (s) = 331.0 ft$
$final velocity (v) = 88.3000 ft/s$

Find

acceleration (a), in ft/s²

Start with the thinking

The governing relation printed in this handbook section is Constant-acceleration kinematics.
Everything except a is given, so isolate a symbolically first — never rearrange after the numbers are in.
Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
Dynamics items reward recognising the unknown before touching a calculator.

Step-by-step solution

Step 1 — State the governing relation:
$v^2 = v_0^2 + 2 a s$
Step 2 — Rearrange the relation so that a stands alone on the left-hand side.
Step 3 — List the givens: initial velocity (v0) = 36.0000 ft/s, distance (s) = 331.0 ft, final velocity (v) = 88.3000 ft/s.
Step 4 — Substitute the given values into the rearranged relation.
Step 5 — Evaluate:
$a = 9.8201\ \text{ft/s²}$
Step 6 — Check: returning a = 9.8201 ft/s² to
$v^2 = v_0^2 + 2 a s$
reproduces the given quantities, and both sides carry the same units.

Answer:

a = 9.8201\ \text{ft/s²}

Why the other options are there

19.6402 — kept a factor of two that cancels in the correct rearrangement.
4.9100 — dropped that same factor in the other direction.
10.8021 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Dynamics → Rigid Body Motion About a Fixed Axis

Example 3

Constant-acceleration kinematics — solve for distance — Rigid Body Motion About a Fixed Axis (3)

A dynamics problem uses Constant-acceleration kinematics. Given initial velocity (v0) = 88.0000 ft/s; acceleration (a) = -17.5000 ft/s²; final velocity (v) = 15.1000 ft/s, determine the distance (s) in ft.

Given

$initial velocity (v_{0}) = 88.0000 ft/s$
$acceleration (a) = -17.5000 ft/s^{2}$
$final velocity (v) = 15.1000 ft/s$

Find

distance (s), in ft

Start with the thinking

The governing relation printed in this handbook section is Constant-acceleration kinematics.
Everything except s is given, so isolate s symbolically first — never rearrange after the numbers are in.
Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
Dynamics items reward recognising the unknown before touching a calculator.

Step-by-step solution

Step 1 — State the governing relation:
$v^2 = v_0^2 + 2 a s$
Step 2 — Rearrange the relation so that s stands alone on the left-hand side.
Step 3 — List the givens: initial velocity (v0) = 88.0000 ft/s, acceleration (a) = -17.5000 ft/s², final velocity (v) = 15.1000 ft/s.
Step 4 — Substitute the given values into the rearranged relation.
Step 5 — Evaluate:
$s = 214.7\ \text{ft}$
Step 6 — Check: returning s = 214.7 ft to
$v^2 = v_0^2 + 2 a s$
reproduces the given quantities, and both sides carry the same units.

Answer:

s = 214.7\ \text{ft}

Why the other options are there

429.5 — kept a factor of two that cancels in the correct rearrangement.
107.4 — dropped that same factor in the other direction.
236.2 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Dynamics → Rigid Body Motion About a Fixed Axis

Example 4

Constant-acceleration kinematics — solve for final velocity (case 2) — Rigid Body Motion About a Fixed Axis (4)

A dynamics problem uses Constant-acceleration kinematics. Given initial velocity (v0) = 33.0000 ft/s; acceleration (a) = 4.5000 ft/s²; distance (s) = 47.0000 ft, determine the final velocity (v) in ft/s.

Given

$initial velocity (v_{0}) = 33.0000 ft/s$
$acceleration (a) = 4.5000 ft/s^{2}$
$distance (s) = 47.0000 ft$

Find

final velocity (v), in ft/s

Start with the thinking

The governing relation printed in this handbook section is Constant-acceleration kinematics.
Everything except v is given, so isolate v symbolically first — never rearrange after the numbers are in.
Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
Dynamics items reward recognising the unknown before touching a calculator.

Step-by-step solution

Step 1 — State the governing relation:
$v^2 = v_0^2 + 2 a s$
Step 2 — Rearrange the relation so that v stands alone on the left-hand side.
Step 3 — List the givens: initial velocity (v0) = 33.0000 ft/s, acceleration (a) = 4.5000 ft/s², distance (s) = 47.0000 ft.
Step 4 — Substitute the given values into the rearranged relation.
Step 5 — Evaluate:
$v = 38.8844\ \text{ft/s}$
Step 6 — Check: returning v = 38.8844 ft/s to
$v^2 = v_0^2 + 2 a s$
reproduces the given quantities, and both sides carry the same units.

Answer:

v = 38.8844\ \text{ft/s}

Why the other options are there

77.7689 — kept a factor of two that cancels in the correct rearrangement.
19.4422 — dropped that same factor in the other direction.
42.7729 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Dynamics → Rigid Body Motion About a Fixed Axis

Example 5

Constant-acceleration kinematics — solve for acceleration (case 2) — Rigid Body Motion About a Fixed Axis (5)

A dynamics problem uses Constant-acceleration kinematics. Given initial velocity (v0) = 89.0000 ft/s; distance (s) = 261.0 ft; final velocity (v) = 23.5000 ft/s, determine the acceleration (a) in ft/s².

Given

$initial velocity (v_{0}) = 89.0000 ft/s$
$distance (s) = 261.0 ft$
$final velocity (v) = 23.5000 ft/s$

Find

acceleration (a), in ft/s²

Start with the thinking

The governing relation printed in this handbook section is Constant-acceleration kinematics.
Everything except a is given, so isolate a symbolically first — never rearrange after the numbers are in.
Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
Dynamics items reward recognising the unknown before touching a calculator.

Step-by-step solution

Step 1 — State the governing relation:
$v^2 = v_0^2 + 2 a s$
Step 2 — Rearrange the relation so that a stands alone on the left-hand side.
Step 3 — List the givens: initial velocity (v0) = 89.0000 ft/s, distance (s) = 261.0 ft, final velocity (v) = 23.5000 ft/s.
Step 4 — Substitute the given values into the rearranged relation.
Step 5 — Evaluate:
$a = -14.1164\ \text{ft/s²}$
Step 6 — Check: returning a = -14.1164 ft/s² to
$v^2 = v_0^2 + 2 a s$
reproduces the given quantities, and both sides carry the same units.

Answer:

a = -14.1164\ \text{ft/s²}

Why the other options are there

-28.2328 — kept a factor of two that cancels in the correct rearrangement.
-7.0582 — dropped that same factor in the other direction.
-15.5280 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Dynamics → Rigid Body Motion About a Fixed Axis

Example 6

Constant-acceleration kinematics — solve for distance (case 2) — Rigid Body Motion About a Fixed Axis (6)

A dynamics problem uses Constant-acceleration kinematics. Given initial velocity (v0) = 75.0000 ft/s; acceleration (a) = 21.0000 ft/s²; final velocity (v) = 24.0000 ft/s, determine the distance (s) in ft.

Given

$initial velocity (v_{0}) = 75.0000 ft/s$
$acceleration (a) = 21.0000 ft/s^{2}$
$final velocity (v) = 24.0000 ft/s$

Find

distance (s), in ft

Start with the thinking

The governing relation printed in this handbook section is Constant-acceleration kinematics.
Everything except s is given, so isolate s symbolically first — never rearrange after the numbers are in.
Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
Dynamics items reward recognising the unknown before touching a calculator.

Step-by-step solution

Step 1 — State the governing relation:
$v^2 = v_0^2 + 2 a s$
Step 2 — Rearrange the relation so that s stands alone on the left-hand side.
Step 3 — List the givens: initial velocity (v0) = 75.0000 ft/s, acceleration (a) = 21.0000 ft/s², final velocity (v) = 24.0000 ft/s.
Step 4 — Substitute the given values into the rearranged relation.
Step 5 — Evaluate:
$s = -120.2\ \text{ft}$
Step 6 — Check: returning s = -120.2 ft to
$v^2 = v_0^2 + 2 a s$
reproduces the given quantities, and both sides carry the same units.

Answer:

s = -120.2\ \text{ft}

Why the other options are there

-240.4 — kept a factor of two that cancels in the correct rearrangement.
-60.1071 — dropped that same factor in the other direction.
-132.2 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Dynamics → Rigid Body Motion About a Fixed Axis

Example 7

Constant-acceleration kinematics — solve for final velocity (case 3) — Rigid Body Motion About a Fixed Axis (7)

A dynamics problem uses Constant-acceleration kinematics. Given initial velocity (v0) = 59.0000 ft/s; acceleration (a) = 11.5000 ft/s²; distance (s) = 218.0 ft, determine the final velocity (v) in ft/s.

Given

$initial velocity (v_{0}) = 59.0000 ft/s$
$acceleration (a) = 11.5000 ft/s^{2}$
$distance (s) = 218.0 ft$

Find

final velocity (v), in ft/s

Start with the thinking

The governing relation printed in this handbook section is Constant-acceleration kinematics.
Everything except v is given, so isolate v symbolically first — never rearrange after the numbers are in.
Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
Dynamics items reward recognising the unknown before touching a calculator.

Step-by-step solution

Step 1 — State the governing relation:
$v^2 = v_0^2 + 2 a s$
Step 2 — Rearrange the relation so that v stands alone on the left-hand side.
Step 3 — List the givens: initial velocity (v0) = 59.0000 ft/s, acceleration (a) = 11.5000 ft/s², distance (s) = 218.0 ft.
Step 4 — Substitute the given values into the rearranged relation.
Step 5 — Evaluate:
$v = 92.1683\ \text{ft/s}$
Step 6 — Check: returning v = 92.1683 ft/s to
$v^2 = v_0^2 + 2 a s$
reproduces the given quantities, and both sides carry the same units.

Answer:

v = 92.1683\ \text{ft/s}

Why the other options are there

184.3 — kept a factor of two that cancels in the correct rearrangement.
46.0842 — dropped that same factor in the other direction.
101.4 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Dynamics → Rigid Body Motion About a Fixed Axis

Example 8

Constant-acceleration kinematics — solve for acceleration (case 3) — Rigid Body Motion About a Fixed Axis (8)

A dynamics problem uses Constant-acceleration kinematics. Given initial velocity (v0) = 94.0000 ft/s; distance (s) = 114.0 ft; final velocity (v) = 146.2 ft/s, determine the acceleration (a) in ft/s².

Given

$initial velocity (v_{0}) = 94.0000 ft/s$
$distance (s) = 114.0 ft$
$final velocity (v) = 146.2 ft/s$

Find

acceleration (a), in ft/s²

Start with the thinking

The governing relation printed in this handbook section is Constant-acceleration kinematics.
Everything except a is given, so isolate a symbolically first — never rearrange after the numbers are in.
Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
Dynamics items reward recognising the unknown before touching a calculator.

Step-by-step solution

Step 1 — State the governing relation:
$v^2 = v_0^2 + 2 a s$
Step 2 — Rearrange the relation so that a stands alone on the left-hand side.
Step 3 — List the givens: initial velocity (v0) = 94.0000 ft/s, distance (s) = 114.0 ft, final velocity (v) = 146.2 ft/s.
Step 4 — Substitute the given values into the rearranged relation.
Step 5 — Evaluate:
$a = 54.9932\ \text{ft/s²}$
Step 6 — Check: returning a = 54.9932 ft/s² to
$v^2 = v_0^2 + 2 a s$
reproduces the given quantities, and both sides carry the same units.

Answer:

a = 54.9932\ \text{ft/s²}

Why the other options are there

110.0 — kept a factor of two that cancels in the correct rearrangement.
27.4966 — dropped that same factor in the other direction.
60.4925 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Dynamics → Rigid Body Motion About a Fixed Axis

Example 9

Constant-acceleration kinematics — solve for distance (case 3) — Rigid Body Motion About a Fixed Axis (9)

A dynamics problem uses Constant-acceleration kinematics. Given initial velocity (v0) = 49.0000 ft/s; acceleration (a) = -9.0000 ft/s²; final velocity (v) = 103.7 ft/s, determine the distance (s) in ft.

Given

$initial velocity (v_{0}) = 49.0000 ft/s$
$acceleration (a) = -9.0000 ft/s^{2}$
$final velocity (v) = 103.7 ft/s$

Find

distance (s), in ft

Start with the thinking

The governing relation printed in this handbook section is Constant-acceleration kinematics.
Everything except s is given, so isolate s symbolically first — never rearrange after the numbers are in.
Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
Dynamics items reward recognising the unknown before touching a calculator.

Step-by-step solution

Step 1 — State the governing relation:
$v^2 = v_0^2 + 2 a s$
Step 2 — Rearrange the relation so that s stands alone on the left-hand side.
Step 3 — List the givens: initial velocity (v0) = 49.0000 ft/s, acceleration (a) = -9.0000 ft/s², final velocity (v) = 103.7 ft/s.
Step 4 — Substitute the given values into the rearranged relation.
Step 5 — Evaluate:
$s = -464.0\ \text{ft}$
Step 6 — Check: returning s = -464.0 ft to
$v^2 = v_0^2 + 2 a s$
reproduces the given quantities, and both sides carry the same units.

Answer:

s = -464.0\ \text{ft}

Why the other options are there

-928.1 — kept a factor of two that cancels in the correct rearrangement.
-232.0 — dropped that same factor in the other direction.
-510.4 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Dynamics → Rigid Body Motion About a Fixed Axis

Example 10

Constant-acceleration kinematics — solve for final velocity (case 4) — Rigid Body Motion About a Fixed Axis (10)

A dynamics problem uses Constant-acceleration kinematics. Given initial velocity (v0) = 23.0000 ft/s; acceleration (a) = 23.0000 ft/s²; distance (s) = 489.0 ft, determine the final velocity (v) in ft/s.

Given

$initial velocity (v_{0}) = 23.0000 ft/s$
$acceleration (a) = 23.0000 ft/s^{2}$
$distance (s) = 489.0 ft$

Find

final velocity (v), in ft/s

Start with the thinking

The governing relation printed in this handbook section is Constant-acceleration kinematics.
Everything except v is given, so isolate v symbolically first — never rearrange after the numbers are in.
Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
Dynamics items reward recognising the unknown before touching a calculator.

Step-by-step solution

Step 1 — State the governing relation:
$v^2 = v_0^2 + 2 a s$
Step 2 — Rearrange the relation so that v stands alone on the left-hand side.
Step 3 — List the givens: initial velocity (v0) = 23.0000 ft/s, acceleration (a) = 23.0000 ft/s², distance (s) = 489.0 ft.
Step 4 — Substitute the given values into the rearranged relation.
Step 5 — Evaluate:
$v = 151.7\ \text{ft/s}$
Step 6 — Check: returning v = 151.7 ft/s to
$v^2 = v_0^2 + 2 a s$
reproduces the given quantities, and both sides carry the same units.

Answer:

v = 151.7\ \text{ft/s}

Why the other options are there

303.5 — kept a factor of two that cancels in the correct rearrangement.
75.8667 — dropped that same factor in the other direction.
166.9 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Dynamics → Rigid Body Motion About a Fixed Axis

Rigid Body Motion About a Fixed Axis

Handbook notes for this section

Core formulas for this FE topic

Worked exam-style examples

Rigid Body Motion About a Fixed Axis

Handbook notes for this section

Core formulas for this FE topic

VariableαConstantα=α0Variable \alpha Constant \alpha = \alpha_{0}VariableαConstantα=α0​

dωω=ω0+α0td\omega \omega = \omega_{0} + \alpha_{0}tdωω=ω0​+α0​t

α=dt\alpha = dtα=dt

ω=dtθ=θ0+ω0t+1α0t2\omega = dt \theta = \theta_{0} + \omega_{0}t + 1 \alpha_{0}t 2ω=dtθ=θ0​+ω0​t+1α0​t2

ω 2 = ω02 + 2α0 _θ - θ0 iNeeds technical review

ωdω=αdθ\omega d\omega = \alpha d\thetaωdω=αdθ

RMq=IqaRMq = Iq aRMq=Iqa

α=Mq/Iq\alpha = Mq /Iqα=Mq/Iq

ω=ω0+αt\omega = \omega_{0} + \alpha tω=ω0​+αt

θ=θ0+ω0t+αt2/2\theta = \theta_{0} + \omega_{0} t + \alpha t_{2}/2θ=θ0​+ω0​t+αt2​/2

Iq ~ 2/2 = Iq ~02 /2 + # MqdiNeeds technical review

Worked exam-style examples