Particle Dynamics - Engineering Mechanics:...
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Particle DynamicsEngineering Mechanics: Dynamics
D. Dane Quinn, PhD
Department of Mechanical Engineering
The University of Akron
Akron OH 44325–3903 USA
Copyright c© 2016
All rights reserved
D. D. Quinn (The University of Akron) Particle Dynamics c© 2016, D. Dane Quinn 1 / 5
Example Projectile Motion
A particle of mass m is launched with an initial inclination θwhile a cart moves with constant velocity
¯vC = u ı and is ini-
tially located a distance d from the particle. Find the initial
speed of the mass so that it lands in the cart. How far must
the ball be thrown?O C
Bv0
θ
d
g
Coordinates and Directions
Only the gravitational force acts on the particle, while the mo-
tion of the particle is in the plane, so that we define x and y to
describe the displacement of the ball as
¯rB/O(t) = x(t) ı + y(t) .
O
B
¯rB/O
y
x
ı
D. D. Quinn (The University of Akron) Particle Dynamics c© 2016, D. Dane Quinn 2 / 5
Example Projectile Motion
The displacement of the cart is defined as
¯rC/O = w ı,
¯vC = w ı = u ı,
so that∫ t
0
{
w(τ ) = u
}
dτ , −→ w(t)− w(0) = u t, (w(0) ≡ d)
w(t) = u t + d.
O C¯rC/O
w
Equations of Motion
The equations of motion are
x = 0, y = −g, with vx = v0 Cθ, vy = v0 Sθ,
x0 = 0, y0 = 0,
so that
x(t) = v0 Cθ, y(t) = −g t + v0 Sθ,
x(t) = v0 Cθ t, y(t) = −g t2
2+ v0 Sθ t.
D. D. Quinn (The University of Akron) Particle Dynamics c© 2016, D. Dane Quinn 3 / 5
Example Projectile Motion
At time t = tf when the ball lands in the cart
¯rB/O(tf ) = ¯
rC/O(tf ), −→ x(tf ) ı + y(tf ) = w(tf ) ı
so that
x(tf ) = w(tf ), y(tf ) = 0,
v0 Cθ tf = d + u tf , −g tf
2
2+ v0 Sθ tf = 0.
From these, we can solve for tf as
tf =2 v0 Sθ
g,
so that
(v0 Cθ−u)2 v0 Sθ
g= d, −→ (Cθ) v0
2 − (u) v0 −d g
2 Sθ= 0.
Finally solving this quadratic equation for the initial speed yields
v0 =u
2 Cθ±
√
(
u
2 Cθ
)2
+d g
2 Sθ Cθ=
u
2 Cθ
{
1±
√
1 +2 d g Cθ
u2 Sθ
}
D. D. Quinn (The University of Akron) Particle Dynamics c© 2016, D. Dane Quinn 4 / 5
Example Projectile Motion
Note that there are two solutions for v0, and only one of them is positive, so that
v0 =u
2 Cθ
{
1 +
√
1 +2 d g Cθ
u2 Sθ
}
.
Therefore, the time and distance at which the mass
lands in the cart is
tf =2 v0 Sθ
g=
u Tθ
g
{
1 +
√
1 +2 d g Cθ
u2 Sθ
}
,
and finally
xf = x(tf ) = w(tf ),
= d +u2 Tθ
g
{
1 +
√
1 +2 d g Cθ
u2 Sθ
}
x
t
y
xf
tf
D. D. Quinn (The University of Akron) Particle Dynamics c© 2016, D. Dane Quinn 5 / 5
