Curvilinear normal and tangential components

1. ES 125

2. Curvilinear Motion: Normal and Tangential
Components.
•When the path along which a particle travels is
known, then it is often convenient to describe
the motion using n and t coordinate axes which
act normal and tangent to the path, respectively,
and at the instant considered have their origin
located at the particle.

3. Curvilinear Motion: Normal and Tangential
Components.

4. Velocity. Since the particle moves, s is a function of time. As indicated in
Sec. 12.4, the particle's velocity v has a direction that is always tangent
to the path, Fig. 12-24c, and a magnitude that is determined by taking
the time derivative of the path function s = s(t),
i.e., v = ds/ dt (Eq. 12-8). Hence

5. Acceleration. The acceleration of the particle
is the time rate of change of the velocity.

6. Acceleration. The acceleration of the particle
is the time rate of change of the velocity.
These two mutually perpendicular
components are shown in Fig. 12-
24 f. Therefore, the magnitude of
acceleration is the positive value
of

7. To better understand these results, consider
the following two special cases of motion.

8. To better understand these results, consider
the following two special cases of motion.

9. Procedure for Analysis

10. Procedure for Analysis

11. Procedure for Analysis

12. EXAMPLE 1:
When the skier reaches
point A along the
parabolic path in Fig. 12-
27a, he has a speed of 6
m/s which is increasing at
2 m/s^2. Determine the
direction of his velocity
and the direction and
magnitude of his
acceleration at this
instant. Neglect the size of
the skier in the calculation.

13. SOLUTION
Coordinate System.
Although the path has
been expressed in terms
of its x and y coordinates,
we can still establish the
origin of the n, t axes at
the fixed point A on the
path and determine the
components of v and a
along these axes, Fig. 12-
27a.

16. EXAMPLE 2:

19. SUPERELEVATION

21. Centrifugal force is the apparent outward
force on a mass when it is rotated.

23. DERIVATION FORMULA OF SUPERELEVATION

24. Ex. 1) A 5.0 kg object moves in a circle at a constant
speed of 10. m/s. What is the radius of the object's
circular path if the object's centripetal force is 1000.
Newtons?

25. Ex. 2) Determine the angle of the super
elevation for a 200 m hi-way curve so that
there will be no slide thrust at a speed of 90
kph.
•A. 19.17°
•B. 17.67°
•C. 18.32°
•D. 20.11°

26. SOLUTION