Projectile Motion-Basic Mechanical Engineering-Lecture Slides, Slides of Mechanical Engineering

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Basic Mechanical Engineering

ME-

Lecture-

Mar 9, 2005

Projectile Motion „

Assumptions

„

No aerodynamic drag

„

No earth’s curvature and rotation effects

„

Small altitude range - g constant

„

a

x

= 0

a

y

= -g

Problem 2/65 S = V

x

• t

t=7.5/10 = .75 secV

f^

= V

y

–gt/

V

y

= 9.81*.75/2= 3.68 m/s

2gh= V

(^2) f

– V

y

2

2

h = 0.689 m

Problem 2/71 S = V

x

* t

800+s*cos40 = 120 * cos40 * t -------(1)t=7.5/10 = .75 sec(y-y

0

) = V

y

t –1/2gt

2

(-ssin40) = 120 * sin40 –1/2gt

2

From (1) & (2)S=1057 m &

t=19.5 sec

Normal and tangential coordinates „

Direction of n +ve towards center of path

Normal and tangential coordinates „

O:is an instantaneous center of rotation

„

:is the instantaneous radius of curvature

„

Advantages:

„

Useful when the velocity sensor is on themoving particle

„

Velocity in the Normal/Tangentialcoordinates has a single component alongthe tangential axes

„

Disadvantages:

„

Moving coordinate systems are inefficientwhen using a fixed monitoring point

Normal and tangential coordinates

For small d

θ

,the change in the instantaneous radius

ρ

may be considered zero,so

Centrifugal acceleration

Normal and tangential coordinates