Normal and Tangential Components

Dynamics · FE Reference Handbook section

Dynamics

3 formulas

10 exam-style examples

~51 min

All Dynamics lectures

Handbook notes for this section

Definitions and conditions exactly as the handbook states them.

Unit vectors et and en are, respectively, tangent and normal to the path with en pointing to the center of curvature. Thus

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 — Normal and Tangential Components

A car travels a curve of radius 25 m at 29 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

$\rho = 25 m$
$v = 29 m/s$
$a_t = 3.0 m/s^{2}$

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
$a_n = 29^{2}/25 = 33.640 m/s^{2}$
Formula
$a = \sqrt(a_t^{2} + a_n^{2})$
Substituting
$a = \sqrt(3.0^{2} + 33.640^{2}) = 33.774 m/s^{2}$
Formula
$\omega = v/\rho$
Substituting
$\omega = 29/25 = 1.1600 rad/s$

Answer:

a_n = 33.64 m/s^{2}, total a = 33.77 m/s^{2}, \omega = 1.160 rad/s

Why the other options are there

36.64 m/s² (components added directly)
21,025 m/s² (multiplied by radius)

Reference: FE Reference Handbook — Dynamics → Normal and Tangential Components

Example 2

Normal and tangential components on a circular path — Normal and Tangential Components (2)

A car travels a curve of radius 170.0 m at 11 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

$\rho = 170.0 m$
$v = 11 m/s$
$a_t = 2.0 m/s^{2}$

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
$a_n = 11^{2}/170.0 = 0.712 m/s^{2}$
Formula
$a = \sqrt(a_t^{2} + a_n^{2})$
Substituting
$a = \sqrt(2.0^{2} + 0.712^{2}) = 2.123 m/s^{2}$
Formula
$\omega = v/\rho$
Substituting
$\omega = 11/170.0 = 0.0647 rad/s$

Answer:

a_n = 0.71 m/s^{2}, total a = 2.12 m/s^{2}, \omega = 0.065 rad/s

Why the other options are there

2.71 m/s² (components added directly)
20,570 m/s² (multiplied by radius)

Reference: FE Reference Handbook — Dynamics → Normal and Tangential Components

Example 3

Normal and tangential components on a circular path — Normal and Tangential Components (3)

A car travels a curve of radius 160.0 m at 16 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

$\rho = 160.0 m$
$v = 16 m/s$
$a_t = 0.5 m/s^{2}$

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
$a_n = 16^{2}/160.0 = 1.600 m/s^{2}$
Formula
$a = \sqrt(a_t^{2} + a_n^{2})$
Substituting
$a = \sqrt(0.5^{2} + 1.600^{2}) = 1.676 m/s^{2}$
Formula
$\omega = v/\rho$
Substituting
$\omega = 16/160.0 = 0.1000 rad/s$

Answer:

a_n = 1.60 m/s^{2}, total a = 1.68 m/s^{2}, \omega = 0.100 rad/s

Why the other options are there

2.10 m/s² (components added directly)
40,960 m/s² (multiplied by radius)

Reference: FE Reference Handbook — Dynamics → Normal and Tangential Components

Example 4

Normal and tangential components on a circular path — Normal and Tangential Components (4)

A car travels a curve of radius 180.0 m at 11 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

$\rho = 180.0 m$
$v = 11 m/s$
$a_t = 1.5 m/s^{2}$

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
$a_n = 11^{2}/180.0 = 0.672 m/s^{2}$
Formula
$a = \sqrt(a_t^{2} + a_n^{2})$
Substituting
$a = \sqrt(1.5^{2} + 0.672^{2}) = 1.644 m/s^{2}$
Formula
$\omega = v/\rho$
Substituting
$\omega = 11/180.0 = 0.0611 rad/s$

Answer:

a_n = 0.67 m/s^{2}, total a = 1.64 m/s^{2}, \omega = 0.061 rad/s

Why the other options are there

2.17 m/s² (components added directly)
21,780 m/s² (multiplied by radius)

Reference: FE Reference Handbook — Dynamics → Normal and Tangential Components

Example 5

Normal and tangential components on a circular path — Normal and Tangential Components (5)

A car travels a curve of radius 85 m at 9 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

$\rho = 85 m$
$v = 9 m/s$
$a_t = 3.0 m/s^{2}$

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
$a_n = 9^{2}/85 = 0.953 m/s^{2}$
Formula
$a = \sqrt(a_t^{2} + a_n^{2})$
Substituting
$a = \sqrt(3.0^{2} + 0.953^{2}) = 3.148 m/s^{2}$
Formula
$\omega = v/\rho$
Substituting
$\omega = 9/85 = 0.1059 rad/s$

Answer:

a_n = 0.95 m/s^{2}, total a = 3.15 m/s^{2}, \omega = 0.106 rad/s

Why the other options are there

3.95 m/s² (components added directly)
6,885 m/s² (multiplied by radius)

Reference: FE Reference Handbook — Dynamics → Normal and Tangential Components

Example 6

Normal and tangential components on a circular path — Normal and Tangential Components (6)

A car travels a curve of radius 195.0 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

$\rho = 195.0 m$
$v = 32 m/s$
$a_t = 2.0 m/s^{2}$

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
$a_n = 32^{2}/195.0 = 5.251 m/s^{2}$
Formula
$a = \sqrt(a_t^{2} + a_n^{2})$
Substituting
$a = \sqrt(2.0^{2} + 5.251^{2}) = 5.619 m/s^{2}$
Formula
$\omega = v/\rho$
Substituting
$\omega = 32/195.0 = 0.1641 rad/s$

Answer:

a_n = 5.25 m/s^{2}, total a = 5.62 m/s^{2}, \omega = 0.164 rad/s

Why the other options are there

7.25 m/s² (components added directly)
199,680 m/s² (multiplied by radius)

Reference: FE Reference Handbook — Dynamics → Normal and Tangential Components

Example 7

Normal and tangential components on a circular path — Normal and Tangential Components (7)

A car travels a curve of radius 180.0 m at 9 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

$\rho = 180.0 m$
$v = 9 m/s$
$a_t = 2.5 m/s^{2}$

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
$a_n = 9^{2}/180.0 = 0.450 m/s^{2}$
Formula
$a = \sqrt(a_t^{2} + a_n^{2})$
Substituting
$a = \sqrt(2.5^{2} + 0.450^{2}) = 2.540 m/s^{2}$
Formula
$\omega = v/\rho$
Substituting
$\omega = 9/180.0 = 0.0500 rad/s$

Answer:

a_n = 0.45 m/s^{2}, total a = 2.54 m/s^{2}, \omega = 0.050 rad/s

Why the other options are there

2.95 m/s² (components added directly)
14,580 m/s² (multiplied by radius)

Reference: FE Reference Handbook — Dynamics → Normal and Tangential Components

Example 8

Normal and tangential components on a circular path — Normal and Tangential Components (8)

A car travels a curve of radius 200.0 m at 29 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

$\rho = 200.0 m$
$v = 29 m/s$
$a_t = 2.5 m/s^{2}$

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
$a_n = 29^{2}/200.0 = 4.205 m/s^{2}$
Formula
$a = \sqrt(a_t^{2} + a_n^{2})$
Substituting
$a = \sqrt(2.5^{2} + 4.205^{2}) = 4.892 m/s^{2}$
Formula
$\omega = v/\rho$
Substituting
$\omega = 29/200.0 = 0.1450 rad/s$

Answer:

a_n = 4.21 m/s^{2}, total a = 4.89 m/s^{2}, \omega = 0.145 rad/s

Why the other options are there

6.71 m/s² (components added directly)
168,200 m/s² (multiplied by radius)

Reference: FE Reference Handbook — Dynamics → Normal and Tangential Components

Example 9

Normal and tangential components on a circular path — Normal and Tangential Components (9)

A car travels a curve of radius 170.0 m at 13 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

$\rho = 170.0 m$
$v = 13 m/s$
$a_t = 1.5 m/s^{2}$

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
$a_n = 13^{2}/170.0 = 0.994 m/s^{2}$
Formula
$a = \sqrt(a_t^{2} + a_n^{2})$
Substituting
$a = \sqrt(1.5^{2} + 0.994^{2}) = 1.800 m/s^{2}$
Formula
$\omega = v/\rho$
Substituting
$\omega = 13/170.0 = 0.0765 rad/s$

Answer:

a_n = 0.99 m/s^{2}, total a = 1.80 m/s^{2}, \omega = 0.076 rad/s

Why the other options are there

2.49 m/s² (components added directly)
28,730 m/s² (multiplied by radius)

Reference: FE Reference Handbook — Dynamics → Normal and Tangential Components

Example 10

Normal and tangential components on a circular path — Normal and Tangential Components (10)

A car travels a curve of radius 30 m at 16 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

$\rho = 30 m$
$v = 16 m/s$
$a_t = 2.0 m/s^{2}$

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
$a_n = 16^{2}/30 = 8.533 m/s^{2}$
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
$\omega = 16/30 = 0.5333 rad/s$

Answer:

a_n = 8.53 m/s^{2}, total a = 8.76 m/s^{2}, \omega = 0.533 rad/s

Why the other options are there

10.53 m/s² (components added directly)
7,680 m/s² (multiplied by radius)

Reference: FE Reference Handbook — Dynamics → Normal and Tangential Components