One of the common descriptions of curvilinear motion uses path variables, which are measurements made along the tangent t and normal n to the path of the particles. t and and n are two orthogonal axes considered separately for every instant of motion.
These coordinates provide a natural description for curvilinear motion and are frequently the most direct and convenient coordinates to use; they move along the path with the particle. The positive direction for n at any position is always taken toward the center of curvature of the path. Therefore no position vector is required.
We now use the coordinates n and t to describe the velocity and acceleration. For this purpose, we introduce unit vectors en in the n-direction and et in the t -direction. During a differential increment of time dt , the particle moves a differential distance ds along the curve from A to A: ds= r d b
where b is in radians.
It is unnecessary to consider the differential change in r between A and A for a differential time period.
The magnitude of velocity v v
ds dt
r
d b b dt
d b dt
r b
The velocity in vector form is v
vet
r b et
The acceleration of the particle is defined as
a
d v
dt
and we observe that the acceleration is a vector which reflects both the change in magnitude and the change in direction of velocity. We now differentiate the velocity applying the ordinary rule for the differentiation of the product of a scalar and a vector and get a
d v
dt
d vet
dt
ve ve t
t
,
The unit vector et now has a nonzero derivative because although its magnitude stays the same, its direction changes. As the particle moves from point A to A ‘, the unit vector et becomes e . t d et
The vectorial difference is: et
et
d et
et
et
et
To find et we analyze the change in during a differential increment of motion as the particle moves from A to A’.
In the limit, the magnitude of d et will be d e e d b d b . For a differential time interval, the direction of d et can be considered perpendicular to . et Therefore, the direction of d et can be considered the same as that of e n . t
t
et
en
d et db
e t
The derivative of et will be obtained as, d et
d b en
d et
or
d b
e
n
By dividing to dt d et
dt
v
b d en dt
e t
r b e
v
t
e t
r b v en
r
,
b e
n
b
v
r
a a
an en
et v
ve t
at et
et v
v
2
r
en
Here, an
v
2
r v
2 r 2 b
r
at
v s r b
v r b
r
v b
( Change in direction of the velocity )
2
an
a
2 r b
r
an
r b 2 v b
2
a2 t
( Change in magnitude of the velocity)
Radius of curvature (Eğrilik Yarıçapı) :
If trajectory is given as y=f(x), the radius of curvature can be determined with the following formula: dy 2 1 dx r
3/ 2
d2y
dx 2
The absolute value sign is used to guarantee that r will always be positive “+”.
The velocity vectors at A and A' are drawn to start at a same point in order to explain the change in the magnitude and direction of the veloicty vector and how these changes generate the normal and tangential components of the acceleration vector.
The change in velocity vector is d v and it indicatesthe direction of the acceleration vector a . When d v is divided into two components one normal and the other tangent to the velocity, the normal component is defined as d v n and in the limit it will be equal to the arc length obtained by rotating the velocity vector with magnitude v for an angle of d b . d v n vd b The normal component of acceleration will be equal to the time derivative of | d v n | .
an
d v n dt
v b
d v
d vt
The tangential component of is . Its magnitude is equal to v v dv or the change in the length of the velocity vector. t
The tangential component of acceleration will be equal to the time derivative of dvt at
dv dt
d vt
dt
v s
The normal component of acceleration an is always directed toward the center of curvature. The tangential component of acceleration at , on the other hand, will be in the positive t -direction of motion if speed v is increasing and in the negative t direction if the speed is decreasing. Figure shows the schematic representations of the variation in the acceleration vector for a particle moving from A to B with (a) increasing speed and (b) decreasing speed.
At an inflection point on the curve, the normal acceleration goes to zero because r becomes infinite. an
v
2
r
v
2
0
Circular motion is an important special case of plane curvilinear motion where the radius of curvature r becomes the constant radius r of the circle and the angle b is replaced by the angle measured from any convenient radial reference to OP.
constant
2
an
at
v r
r = r =
v
d dt d dt
r
r 2 v
(Angular velocity)
(Angular acceleration)
1. Six acceleration vectors are shown for the car whose velocity vector is directed forward. For each acceleration vector describe in words the instantaneous motion of the car.
A flywheel rotates with variable velocity. At a certain time, the tangential acceleration of point A on the flywheel is 1 m/s2, while the normal acceleration of point B on the flywheel is 0.6 m/s 2. Determine the velocity of point A and the total acceleration of point B. 2.
200 mm
A B 150 mm
A baseball player releases a ball with the initial conditions shown in the figure. Determine the radius of curvature of the trajectory (a) just after release and (b) at the apex. For each case, compute the time rate of change of the speed. 3.
A marble rolls down a chute which is bent in the shape of a parabola. Its equation is given as 4.
y=x2 – 6x + 9
[m]
The marble passes point A with a speed of 3 m/s, which is increasing at the rate of 5 m/s2. Determine the normal and tangential components an and at , of the acceleration of the marble as it passes point A. Also determine the angle between the velocity vector and the acceleration vector at point A.
y (m) y=f(x) A
x=5 m
x (m)
5. A ball is thrown horizontally from the top of a 50 m cliff at A with a speed of 15 m/s and lands at point C. Because of strong horizontal wind the ball has a constant acceleration in the negative x-direction. Determine the radius of curvature r of the path of the ball at B where its trajectory makes an angle of 45o with the horizontal. Neglect any effect of air resistance in the vertical direction.
6. During a short interval, the slotted guides are designed to move according to x = 16 – 12t + 4t 2 and y = 2 + 15t – 3t 2, where x and y are in millimeters and t is in seconds. At the instant when t = 2s, determine the radius of curvature r of the path of the constrained pin P.
Homework Pin P in the crank PO engages the horizontal slot in the guide C and controls its motion on the fixed vertical rod. Determine the velocity y and the acceleration yof guide C for a given value of the angle is 0 (a) and (b) if =0 and
