One of the common descriptions of curvilinear measurements ... · One of the common descriptions of...
Transcript of One of the common descriptions of curvilinear measurements ... · One of the common descriptions of...
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 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 in the n-direction and in the t-direction.
ne
te
During a differential increment of time dt, the particle moves a differential distance ds along the curve from A to A:
ds=r db
where b is in radians.
It is unnecessary to consider the differential change in r
between A and A for a differential time period.
brb
r dt
d
dt
dsv
tev
br
The magnitude of velocity:
br
bbb
r
v
dt
d
dt
d
dt
dsv
Therefore, the velocity in vector form is
tt eevv
br
The acceleration of the particle is defined as ,
and we observe that the acceleration is a vector which reflects both the change in magnitude and the change in direction of velocity.
dt
vda
We now differentiate the velocity applying the ordinary rule for the differentiation of the product of a scalar and a vector and get
tt
t evevdt
evd
dt
vda
The unit vector 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 becomes .
te
te
te
te
te
ted
ttt edee
The vectorial difference is:
To find we analyze the change in during a differential
increment of motion as the particle moves from A to A’.
te
te
In the limit, the magnitude of will
be . For a differential
time interval, the direction of can
be considered perpendicular to .
Therefore, the direction of can
be considered the same as that of .
ted
bb ddeed tt
te
ted
ted
ne
db
ted
te
te
ne
The derivative of will be obtained as, te
nt
nt ed
edoeded
bb r
By dividing to dt
nt
t
ntnt
ev
e
vvev
eeedt
d
dt
ed
r
rbbrbr
bb
,
nttt
ttnn
ev
eveveva
eaeaa
r
2
nttt
n
a
t
a
tt
ev
veveveva
veveveveva
nt
r
rbbrb
v
222
a ntnttt aaev
eveveva
r
Here,
22
22
2222
tn
t
n
n
aaa
sva
vv
a
vvv
a
br
bbrr
br
brbr
r
br
r
(Change in direction of the velocity)
(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:
The absolute value sign is used to guarantee that r will always be positive “+”.
2
2
2/32
1
dx
yd
dx
dy
r
The change in velocity vector is and it indicates the direction of the acceleration vector . When is divided into two components one normal and the other tangent to the velocity, the normal component is defined as 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 db. The normal component of acceleration will be equal to the time derivative of | | .
vd
a
vd
bvdvd n
b
vdt
vda
n
n
nvd
nvd
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 tangential component of is . Its magnitude is equal to or the change in the length of the velocity vector. The tangential component of acceleration will be equal to the time derivative of
vd
tvd
tdvvv
tdv
svdt
vd
dt
dva
t
t
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.
022
r
vv
an
At an inflection point on the curve, the normal acceleration goes to zero since r becomes infinite.
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.
r = r = constant
rva
vrr
va
t
n
22
dt
d
dt
d
(Angular velocity)
(Angular acceleration)
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.
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.
2. 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/s2. Determine the velocity of point A and the total acceleration of point B.
A
B 150 mm
200 mm
3. 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.
4. A marble rolls down a chute which is bent in the shape of a parabola. Its equation is given as 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.
x (m)
y (m)
x=5 m
A
y=f(x)
5. The pin P is constrained to move in the slotted guides which move at right angles to one another. At the instant represented, A has a velocity to the right of 0.2 m/s which is decreasing at the rate of 0.75 m/s each second. At the same time, B is moving down with a velocity of 0.15 m/s which is decreasing at the rate of 0.5 m/s each second. For this instant, determine the radius of curvature r of the path followed by P.
6. 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.
7. In the design of a control mechanism, the vertical slotted guide is moving with a constant velocity = 150 mm/s during the interval of its motion from x = -80 mm to x = +80 mm. For the instant when x = 60 mm, calculate the n- and t-components of acceleration of the pin P, which is confined to move in the parabolic groove. From these results, determine the radius of curvature r of the path at this position.
x