Dependent Motion

Dependent Motion
• Position of a particle may depend on position of one
or more other particles.
Engr 240 – Week 2
• Position of block B depends on position of block A.
Since rope is of constant length, it follows that sum of
lengths of segments must be constant.
x A + 2 x B = constant (one degree of freedom)
• Positions of three blocks are dependent.
Dependent Motion, Graphical Solution,
Curvilinear Motion
2 x A + 2 x B + xC = constant (two degrees of freedom)
• For linearly related positions, similar relations hold
between velocities and accelerations.
dx
dx A
dx
+ 2 B + C = 0 or 2v A + 2v B + vC = 0
dt
dt
dt
dv
dv
dv
2 A + 2 B + C = 0 or 2a A + 2a B + aC = 0
dt
dt
dt
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Sample Problem 11.5
SOLUTION:
• Define origin at upper horizontal surface
with positive displacement downward.
Pulley D is attached to a collar which
is pulled down at 3 in./s. At t = 0,
collar A starts moving down from K
with constant acceleration and zero
initial velocity. Knowing that
velocity of collar A is 12 in./s as it
passes L, determine the change in
elevation, velocity, and acceleration
of block B when block A is at L.
Graphical Solution of Rectilinear-Motion
Problems
Graphical Solution of Rectilinear-Motion
Problems
• Given the x-t curve, the v-t curve is
equal to the x-t curve slope.
• Given the a-t curve, the change in velocity between t1 and
t2 is equal to the area under the a-t curve between t1 and t2.
• Given the v-t curve, the a-t curve is
equal to the v-t curve slope.
• Given the v-t curve, the change in position between t1 and
t2 is equal to the area under the v-t curve between t1 and t2.
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Example.
A subway train travels between
two stations with the
acceleration schedule shown.
Find the time interval Δt during
which the acceleration is 2
m/s2, and the distance s
between the stations.
CURVILINEAR MOTION OF PARTICLES
– Motion along a curve other than a straight line
• Position vector - the vector between origin
O of reference frame and the position
occupied by particle.
• Consider particle which occupies position P defined
r
r
by r at time t and P’ defined by r ′ at t + Δt,
r
r
Δr dr
r
v = lim
=
dt
Δt →0 Δt
= instantaneous velocity (vector)
v = lim
Δs
Δt →0 Δt
=
ds
dt
= instantaneous speed (scalar)
r
• Consider velocity v of particle at time t and velocity
r
v′ at t + Δt,
r
r
Δv dv
r
=
a = lim
dt
Δt →0 Δt
= instantaneous acceleration (vector)
• In general, acceleration vector is not tangent to
particle path and velocity vector.
Rectangular Components
r
r = x iˆ + y ˆj + z kˆ
r
r dr
= x& iˆ + y& ˆj + z& kˆ
v=
dt
r
r dv
= &x& iˆ + &y& ˆj + &z& kˆ
a=
dt
where the “overdots” represent
derivatives with respect to time.
Example: Projectile Motion
(x-y plane along the plane of motion)
x: uniform motion
y: free fall
Equations of Motion:
z: no motion
Example.
A projectile is fired from the edge of
a 150-m cliff with an initial
velocity of 180 m/s at an angle
of 30° with the horizontal.
Neglecting air resistance, find
a) the horizontal distance from the
gun to the point where the
projectile strikes the ground,
b) the greatest elevation above the
ground reached by the
projectile.
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Normal and Tangential Components
• Velocity vector of particle is tangent to path of
particle.
r
v = veˆt
r
r
• et and et′ are tangential unit vectors for the
particle path at P and P’.
r r r
Δet = et′ − et
• If Δθ is the angle between the two unit vectors,
Δet = 2 sin (Δθ 2 )
r
Δet
sin (Δθ 2 ) r
r
en = en
lim
= lim
Δθ →0 Δθ
Δθ →0 Δθ 2
r
de
r
en = t
dθ
Acceleration
r
v = veˆt
• Differential velocity vector to find acceleration.
r
r
r
de dθ ds
de dv r
r dv dv r
= et + v
= et + v
a=
dθ ds dt
dt dt
dt dt
but r
det r
ds
= en
ρ dθ = ds
=v
dθ
dt
After substituting,
dv
v2
r dv r v 2 r
a = et + en
at =
an =
ρ
ρ
dt
dt
• Tangential Component of acceleration – change
in magnitude of velocity vector .
• Normal Component – change in direction of
velocity vector; towards center of curvature.
Example (Problem 11.10)
A motorist is traveling on a curved section of highway of radius
2500 ft at the speed of 60 mi/h. The motorist suddenly applies
the brakes, causing the automobile to slow down at a constant
rate. Knowing that after 8 s the speed has been reduced to 45
mi/h, determine the acceleration of the automobile immediately
after the brakes have been applied.
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