Plane Circular Motion

Dynamics · FE Reference Handbook section

Dynamics

11 formulas

10 exam-style examples

~60 min

All Dynamics lectures

Handbook notes for this section

Definitions and conditions exactly as the handbook states them.

A special case of radial and transverse components is for constant radius rotation about the origin, or plane circular motion.
Here the vector quantities are defined as
The values of the angular velocity and acceleration, respectively, are defined as
Arc length, transverse velocity, and transverse acceleration, respectively, are
The radial acceleration is given by

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

Normal and tangential components on a circular path — Plane Circular Motion

A car travels a curve of radius 120.0 m at 32 m/s while speeding up at 2.0 m/s². Resolve the acceleration into normal and tangential components, find the total acceleration, and compute the angular velocity of the radius vector.

Given

$\rho = 120.0 m$
$v = 32 m/s$
$a_t = 2.0 m/s^{2}$

Find

a_n, total a, and ω

Start with the thinking

The tangential component changes speed; the normal component changes direction and always points to the centre.
Total acceleration is the vector sum of the two perpendicular components.

Step-by-step solution

Formula
$a_n = v^{2}/\rho$
Substituting
$a_n = 32^{2}/120.0 = 8.533 m/s^{2}$
Formula
$a = \sqrt(a_t^{2} + a_n^{2})$
Substituting
$a = \sqrt(2.0^{2} + 8.533^{2}) = 8.765 m/s^{2}$
Formula
$\omega = v/\rho$
Substituting
$\omega = 32/120.0 = 0.2667 rad/s$

Answer:

a_n = 8.53 m/s^{2}, total a = 8.76 m/s^{2}, \omega = 0.267 rad/s

Why the other options are there

10.53 m/s² (components added directly)
122,880 m/s² (multiplied by radius)

Reference: FE Reference Handbook — Dynamics → Plane Circular Motion

Example 2

Normal and tangential components on a circular path — Plane Circular Motion (2)

A car travels a curve of radius 180.0 m at 30 m/s while speeding up at 1.0 m/s². Resolve the acceleration into normal and tangential components, find the total acceleration, and compute the angular velocity of the radius vector.

Given

$\rho = 180.0 m$
$v = 30 m/s$
$a_t = 1.0 m/s^{2}$

Find

a_n, total a, and ω

Start with the thinking

The tangential component changes speed; the normal component changes direction and always points to the centre.
Total acceleration is the vector sum of the two perpendicular components.

Step-by-step solution

Formula
$a_n = v^{2}/\rho$
Substituting
$a_n = 30^{2}/180.0 = 5.000 m/s^{2}$
Formula
$a = \sqrt(a_t^{2} + a_n^{2})$
Substituting
$a = \sqrt(1.0^{2} + 5.000^{2}) = 5.099 m/s^{2}$
Formula
$\omega = v/\rho$
Substituting
$\omega = 30/180.0 = 0.1667 rad/s$

Answer:

a_n = 5.00 m/s^{2}, total a = 5.10 m/s^{2}, \omega = 0.167 rad/s

Why the other options are there

6.00 m/s² (components added directly)
162,000 m/s² (multiplied by radius)

Reference: FE Reference Handbook — Dynamics → Plane Circular Motion

Example 3

Normal and tangential components on a circular path — Plane Circular Motion (3)

A car travels a curve of radius 165.0 m at 15 m/s while speeding up at 1.5 m/s². Resolve the acceleration into normal and tangential components, find the total acceleration, and compute the angular velocity of the radius vector.

Given

$\rho = 165.0 m$
$v = 15 m/s$
$a_t = 1.5 m/s^{2}$

Find

a_n, total a, and ω

Start with the thinking

The tangential component changes speed; the normal component changes direction and always points to the centre.
Total acceleration is the vector sum of the two perpendicular components.

Step-by-step solution

Formula
$a_n = v^{2}/\rho$
Substituting
$a_n = 15^{2}/165.0 = 1.364 m/s^{2}$
Formula
$a = \sqrt(a_t^{2} + a_n^{2})$
Substituting
$a = \sqrt(1.5^{2} + 1.364^{2}) = 2.027 m/s^{2}$
Formula
$\omega = v/\rho$
Substituting
$\omega = 15/165.0 = 0.0909 rad/s$

Answer:

a_n = 1.36 m/s^{2}, total a = 2.03 m/s^{2}, \omega = 0.091 rad/s

Why the other options are there

2.86 m/s² (components added directly)
37,125 m/s² (multiplied by radius)

Reference: FE Reference Handbook — Dynamics → Plane Circular Motion

Example 4

Normal and tangential components on a circular path — Plane Circular Motion (4)

A car travels a curve of radius 135.0 m at 8 m/s while speeding up at 1.5 m/s². Resolve the acceleration into normal and tangential components, find the total acceleration, and compute the angular velocity of the radius vector.

Given

$\rho = 135.0 m$
$v = 8 m/s$
$a_t = 1.5 m/s^{2}$

Find

a_n, total a, and ω

Start with the thinking

The tangential component changes speed; the normal component changes direction and always points to the centre.
Total acceleration is the vector sum of the two perpendicular components.

Step-by-step solution

Formula
$a_n = v^{2}/\rho$
Substituting
$a_n = 8^{2}/135.0 = 0.474 m/s^{2}$
Formula
$a = \sqrt(a_t^{2} + a_n^{2})$
Substituting
$a = \sqrt(1.5^{2} + 0.474^{2}) = 1.573 m/s^{2}$
Formula
$\omega = v/\rho$
Substituting
$\omega = 8/135.0 = 0.0593 rad/s$

Answer:

a_n = 0.47 m/s^{2}, total a = 1.57 m/s^{2}, \omega = 0.059 rad/s

Why the other options are there

1.97 m/s² (components added directly)
8,640 m/s² (multiplied by radius)

Reference: FE Reference Handbook — Dynamics → Plane Circular Motion

Example 5

Normal and tangential components on a circular path — Plane Circular Motion (5)

A car travels a curve of radius 155.0 m at 17 m/s while speeding up at 1.0 m/s². Resolve the acceleration into normal and tangential components, find the total acceleration, and compute the angular velocity of the radius vector.

Given

$\rho = 155.0 m$
$v = 17 m/s$
$a_t = 1.0 m/s^{2}$

Find

a_n, total a, and ω

Start with the thinking

The tangential component changes speed; the normal component changes direction and always points to the centre.
Total acceleration is the vector sum of the two perpendicular components.

Step-by-step solution

Formula
$a_n = v^{2}/\rho$
Substituting
$a_n = 17^{2}/155.0 = 1.865 m/s^{2}$
Formula
$a = \sqrt(a_t^{2} + a_n^{2})$
Substituting
$a = \sqrt(1.0^{2} + 1.865^{2}) = 2.116 m/s^{2}$
Formula
$\omega = v/\rho$
Substituting
$\omega = 17/155.0 = 0.1097 rad/s$

Answer:

a_n = 1.86 m/s^{2}, total a = 2.12 m/s^{2}, \omega = 0.110 rad/s

Why the other options are there

2.86 m/s² (components added directly)
44,795 m/s² (multiplied by radius)

Reference: FE Reference Handbook — Dynamics → Plane Circular Motion

Example 6

Normal and tangential components on a circular path — Plane Circular Motion (6)

A car travels a curve of radius 100.0 m at 19 m/s while speeding up at 2.5 m/s². Resolve the acceleration into normal and tangential components, find the total acceleration, and compute the angular velocity of the radius vector.

Given

$\rho = 100.0 m$
$v = 19 m/s$
$a_t = 2.5 m/s^{2}$

Find

a_n, total a, and ω

Start with the thinking

The tangential component changes speed; the normal component changes direction and always points to the centre.
Total acceleration is the vector sum of the two perpendicular components.

Step-by-step solution

Formula
$a_n = v^{2}/\rho$
Substituting
$a_n = 19^{2}/100.0 = 3.610 m/s^{2}$
Formula
$a = \sqrt(a_t^{2} + a_n^{2})$
Substituting
$a = \sqrt(2.5^{2} + 3.610^{2}) = 4.391 m/s^{2}$
Formula
$\omega = v/\rho$
Substituting
$\omega = 19/100.0 = 0.1900 rad/s$

Answer:

a_n = 3.61 m/s^{2}, total a = 4.39 m/s^{2}, \omega = 0.190 rad/s

Why the other options are there

6.11 m/s² (components added directly)
36,100 m/s² (multiplied by radius)

Reference: FE Reference Handbook — Dynamics → Plane Circular Motion

Example 7

Normal and tangential components on a circular path — Plane Circular Motion (7)

A car travels a curve of radius 110.0 m at 15 m/s while speeding up at 3.0 m/s². Resolve the acceleration into normal and tangential components, find the total acceleration, and compute the angular velocity of the radius vector.

Given

$\rho = 110.0 m$
$v = 15 m/s$
$a_t = 3.0 m/s^{2}$

Find

a_n, total a, and ω

Start with the thinking

The tangential component changes speed; the normal component changes direction and always points to the centre.
Total acceleration is the vector sum of the two perpendicular components.

Step-by-step solution

Formula
$a_n = v^{2}/\rho$
Substituting
$a_n = 15^{2}/110.0 = 2.045 m/s^{2}$
Formula
$a = \sqrt(a_t^{2} + a_n^{2})$
Substituting
$a = \sqrt(3.0^{2} + 2.045^{2}) = 3.631 m/s^{2}$
Formula
$\omega = v/\rho$
Substituting
$\omega = 15/110.0 = 0.1364 rad/s$

Answer:

a_n = 2.05 m/s^{2}, total a = 3.63 m/s^{2}, \omega = 0.136 rad/s

Why the other options are there

5.05 m/s² (components added directly)
24,750 m/s² (multiplied by radius)

Reference: FE Reference Handbook — Dynamics → Plane Circular Motion

Example 8

Normal and tangential components on a circular path — Plane Circular Motion (8)

A car travels a curve of radius 100.0 m at 25 m/s while speeding up at 1.5 m/s². Resolve the acceleration into normal and tangential components, find the total acceleration, and compute the angular velocity of the radius vector.

Given

$\rho = 100.0 m$
$v = 25 m/s$
$a_t = 1.5 m/s^{2}$

Find

a_n, total a, and ω

Start with the thinking

The tangential component changes speed; the normal component changes direction and always points to the centre.
Total acceleration is the vector sum of the two perpendicular components.

Step-by-step solution

Formula
$a_n = v^{2}/\rho$
Substituting
$a_n = 25^{2}/100.0 = 6.250 m/s^{2}$
Formula
$a = \sqrt(a_t^{2} + a_n^{2})$
Substituting
$a = \sqrt(1.5^{2} + 6.250^{2}) = 6.427 m/s^{2}$
Formula
$\omega = v/\rho$
Substituting
$\omega = 25/100.0 = 0.2500 rad/s$

Answer:

a_n = 6.25 m/s^{2}, total a = 6.43 m/s^{2}, \omega = 0.250 rad/s

Why the other options are there

7.75 m/s² (components added directly)
62,500 m/s² (multiplied by radius)

Reference: FE Reference Handbook — Dynamics → Plane Circular Motion

Example 9

Normal and tangential components on a circular path — Plane Circular Motion (9)

A car travels a curve of radius 120.0 m at 29 m/s while speeding up at 0.5 m/s². Resolve the acceleration into normal and tangential components, find the total acceleration, and compute the angular velocity of the radius vector.

Given

$\rho = 120.0 m$
$v = 29 m/s$
$a_t = 0.5 m/s^{2}$

Find

a_n, total a, and ω

Start with the thinking

The tangential component changes speed; the normal component changes direction and always points to the centre.
Total acceleration is the vector sum of the two perpendicular components.

Step-by-step solution

Formula
$a_n = v^{2}/\rho$
Substituting
$a_n = 29^{2}/120.0 = 7.008 m/s^{2}$
Formula
$a = \sqrt(a_t^{2} + a_n^{2})$
Substituting
$a = \sqrt(0.5^{2} + 7.008^{2}) = 7.026 m/s^{2}$
Formula
$\omega = v/\rho$
Substituting
$\omega = 29/120.0 = 0.2417 rad/s$

Answer:

a_n = 7.01 m/s^{2}, total a = 7.03 m/s^{2}, \omega = 0.242 rad/s

Why the other options are there

7.51 m/s² (components added directly)
100,920 m/s² (multiplied by radius)

Reference: FE Reference Handbook — Dynamics → Plane Circular Motion

Example 10

Normal and tangential components on a circular path — Plane Circular Motion (10)

A car travels a curve of radius 190.0 m at 21 m/s while speeding up at 3.0 m/s². Resolve the acceleration into normal and tangential components, find the total acceleration, and compute the angular velocity of the radius vector.

Given

$\rho = 190.0 m$
$v = 21 m/s$
$a_t = 3.0 m/s^{2}$

Find

a_n, total a, and ω

Start with the thinking

The tangential component changes speed; the normal component changes direction and always points to the centre.
Total acceleration is the vector sum of the two perpendicular components.

Step-by-step solution

Formula
$a_n = v^{2}/\rho$
Substituting
$a_n = 21^{2}/190.0 = 2.321 m/s^{2}$
Formula
$a = \sqrt(a_t^{2} + a_n^{2})$
Substituting
$a = \sqrt(3.0^{2} + 2.321^{2}) = 3.793 m/s^{2}$
Formula
$\omega = v/\rho$
Substituting
$\omega = 21/190.0 = 0.1105 rad/s$

Answer:

a_n = 2.32 m/s^{2}, total a = 3.79 m/s^{2}, \omega = 0.111 rad/s

Why the other options are there

5.32 m/s² (components added directly)
83,790 m/s² (multiplied by radius)

Reference: FE Reference Handbook — Dynamics → Plane Circular Motion