LESSON 6

TANGENTIAL AND NORMAL COMPONENTS OF

ACCELERATION

Overview:
This lesson covers components of acceleration. It can be broken into
tangential and normal components; the tangential vector points in the direction in
which an object is moving and the normal vector points in the direction in which
the curve of that object's path is turning.

Learning Outcome:
At the end of this lesson, the students can compute normal and tangential
components of acceleration

Materials Needed:
Hand-outs, Chalkboard/chalk, Eraser, Laptop, Projector and PPT
presentation

Duration: 4 hours

Learning Content:
TANGENTIAL AND NORMAL COMPONENTS OF ACCELERATION

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at = tangential acceleration
dV
at = (rate of change in magnitude of velocity and it will be zero if the speed is
dt

constant)
ax Vx ay Vy ax Vx + ay Vy
at = ax Cos θ + ay Sin θ = + =
V V V

an = normal acceleration
V2
an =
r
V = rω
r2 ω2
an = = rω2
r
ω = angular speed in rad/sec
V = velocity at any point which is tangent to the path
r = radius of curvature at any point A.
a x Vx a y Vy ax Vx − ay Vy
an = ay Cos θ – ax Sin θ = - =
V V V

V = √(Vx )2 + (Vy )2

a = √(ax )2 + (ay )2 = √(at )2 + (an )2

Example No. 1
At the bottom of the loop, the speed of an airplane is 400 mph. This causes a
normal acceleration of 9g ft/sec2. Determine the radius of the loop.
Solution:
V = 400 miles/hr (5280 ft/mile) (1 hr/3600 sec) = 586.76 fps
(586.67)2
9(32.2) =
r
r = 1,188 ft.

Example No. 2
A particle moves on a circular path of 20 ft radius so that its arc distance from a
fixed point on the path is given by S = 4t 3 – 10t where S is in feet and t in
seconds.
A. Compute the tangential acceleration at the end of 2 seconds.
B. Compute the normal acceleration at the end of 2 seconds.
C. Compute the total acceleration at the end of 2 seconds.

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Solution:
A. Tangential acceleration
S = 4t3 – 10t
dS
= 12t2 -10 = 12(2)2 – 10 = 38 fps
dt
V = 38 fps

V = 12t2 -10
dV
= 24t = 24(2) = 48 fps2
dt
at = 48 fps2

B. Normal acceleration
V2 (38)2
an = = = 72.2 fps2
r 20
C. Total acceleration
a = √(at )2 + (an )2 = √(48)2 + (72.2)2 = 86.7 fps2

Example No. 3
A particle is moving along a curved path. At a certain instant when the slope of
the path is 0.75, ax = 6 fps2 and ay = 10 fps2. Compute
A. the normal acceleration
B. the total acceleration
C. the tangential acceleration

Solution:
A. Normal acceleration
tan θ = 0.75
θ = 36.870
an = ay Cos θ – ax Sin θ

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 an = 10 Cos 36.870 – 6 Sin 36.870
an = 4.4 m/s2

B. Total acceleration

a = √(ax )2 + (ay )2

a = √(6)2 + (10)2
a = 11.66 m/s2

C. Tangential acceleration

a = √(at )2 + (an )2

11.66 = √(at )2 + (4.4)2

at = 10.8 m/s2

Learning Activity:
Seatwork
1. A stone is thrown with an initial velocity of 100 fps upward at 60 0 to the
horizontal. Compute
A). the velocity of the stone when it is 50 feet horizontally from its initial
position.
B). the normal acceleration when it is 50 feet horizontally from its initial
position.
C). the radius of curvature of its path where it is 50 feet horizontally from
its position.

Ans. A). y = 70.50 ft

Vy = 54.40 fps
Vx = 50 ft
V = 73.89 fps

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 B). α = 47.410
ax = 0
ay = 32.2 fps2
an = 21.79 fps2
C). r = 250.56 ft

2. A particle moves on a circle in accordance with the equation S = t 4 – 8t where

S is the displacement in feet measured along the circular path and t is in
seconds after starting from rest, the total acceleration of the particle is 48√2
fps2.
A). Compute the velocity of the particle after 2 seconds
B). Compute the normal acceleration of the particle after 2 seconds.
C). Compute the radius of the circle.

Ans. A). V = 24 fps

B). an = 48 fps2
C). r =12 ft

3. A motorcycle starts from rest at t = 0 on a circular track of 400 m radius. The

tangential component of the motorcycle’s acceleration is at = 2 + 0.2t m/s2
A). Determine the velocity of the motorcycle at t = 10 sec.
B). determine the distance the motorcycle has moved along the circular
track at t =10 seconds.
C). Determine the magnitude of its acceleration at t = 10 seconds.

Ans. A). V = 30 m/s

B). S = 133.33 m
C). at = 4 m/s2
an = 2.25 m/s2
a = 4.59 m/s2

Learning Evaluation:
Solve the following problems neatly. Show all your solutions:

1. A particle P moves in a circular path shown. Determine the magnitude of the

acceleration when

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 A). the speed V = 1.2 m/s is constant
B). the speed is 1.2 m/s and is increasing at a rate of 2.4 m/s each
second.
C). the speed is 1.2 m/s and is decreasing at the rate of 4.8 m/s each
second.

Ans. A). at = 0 when V is constant

an = 2.4 m/s2
a = 2.4 m/s2
B). at = 2.4 m/s2
an = 2.4 m/s2
a = 3.39 m/s2
C). at = - 4.8 m/s2
an = 2.4 m/s2
a = 5.37 m/s2

2. A race driver travelling at a speed of 250 kph on a straight highway applies his
bakes at point A and reduce his speed at a uniform rate of 200 kph at point C
in a distance of 300 m.
A). Calculate the velocity at point B
B). Calculate the normal acceleration at point B.
C). Calculate the magnitude of the total acceleration of the race car an
instant after it passes point B.

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Ans. A). at = - 8.04 m/s2
VB = 49.1 m/s
B). an = 4.82 m/s2
C). a = 9.37 m/s2

3. A car travels along the level curved road with an acceleration that is
decreasing at the constant rate of 0.6 m/s2. The speed of the car as it passes
point A is 16 m/s. Radius of curvature of the road at point B is 60 m.
A). Compute the velocity of the car at point B which is 120 m along the
road from point A
B). Compute the time of travel of the car from A to B.
C). Compute the magnitude of the total acceleration of the car as it passes
point B which is 120 m along the road from A.

Ans. A). VB = 10.58 m/s

B). t = 9.03 sec
C). an = 1.87 m/s2
at = - 0.6 m/s2
a = 1.96 m/s2

4. The Cartesian coordinates of a point in meters are:

x = 2t + 4 y = t3 – 3t z = 2t2 + 4
A). What is the magnitude of the velocity of the point at t = 2 sec?
B). What is the magnitude of the acceleration of the point at t = 2 sec?
C). What are the coordinates of the point when t = 2 sec?

Ans. A). Vx = 2 m/s

Vy = 9 m/s
Vz = 8 m/s
V = 12.21 m/s
B). ax = 0

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 ay = 12 m/s2
az = 4 m/s2
a = 12.65 m/s2
C). Pt (8, 2, 12)

References:
Arreola, M. A. (1996). Engineering Mechanics. KEN, Inc.
Besavilla, V. I. (1986). Engineering Mechanics. 2nd edition. VIB Publisher
Estanero, et. al. (2008). Principles of Engineering Mechanics: Statics. C and E
Publishing, DLSU – Manila
Singer, F. L (1980). Engineering Mechanics. 3nd edition. New York. Harper and
Row
Timoshenko, S & Young, D. H. (1956). Engineering Mechanics. Tokyo. McGraw-
Hill

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