Problem 5/125 Page 384, Engineering Mechanics - Dynamics, Meriam and Kraige, 4th Edition
Given: A bar of length L with right end moving a constant velocity of V0 to the right and left end moving along a circle of radius R. At the instant of interest the left of the bar is a distance H above the right end of the bar. The direction of motion of the left end of the bar makes an angle q with the horizontal
Find: The angular acceleration of the bar and the tangent acceleration of the left end of the bar.
L = 1.2 m
R = 0.5 m
V0 = 3 m/s
H = 0.25 m
q = 60 degrees
The accelerations of the two ends of the bar are related by the acceleration
two point formula. Considering +X to be horizontal to the right and
+Y to be vertically upward:
aRE = aLE + w X ( w
X r ) + a X r
The right end is moving with a constant velocity and thus has zero acceleration.
0 = aLE + w k X
( w k X ( (L2 -
H2)1/2 i - H j ) ) + a
k X ( (L2 - H2)1/2
i - H j )
The left end is moving along a curved path and thus has both a tangent
acceleration and a normal acceleration. The tangent acceleration
is parallel to the path and thus makes an angle of q
with the horizontal. The normal acceleration is perpendicular to
the path and directed toward the center of curvature. Thus the normal
acceleration makes an angle of 90 degrees - q
with the horizontal (note that the relationships between sines and cosines
of 90 degrees minus an angle are exploited in the following). The
magnitude of the normal acceleration is the square of the speed of the
left end divided by the radius of curvature of the path. Using these
facts and performing the required cross products:
0 = aLEt ( cos q i -
sin q j ) + vLE2/R
( sin q i + cos q
j ) + w k X ( w(L2
- H2)1/2 j + w H
i ) + a(L2 - H2)1/2
j + a H i
0 = aLEt ( cos q i -
sin q j ) + vLE2/R
( sin q i + cos q
j ) - w2(L2
- H2)1/2 i + w2
H j + a(L2
- H2)1/2 j + a H
i
Equating components:
0 = aLEt cos q + vLE2/R
sin q - w2(L2
- H2)1/2 + a H
0 = -aLEt sin q
+ vLE2/R cos q + w2
H + a(L2 - H2)1/2
Noting that the speed of the left end and the angular speed of the bar
are known from problem 5-63, the above provides two equations in the two
unknowns of interest, the tangent acceleration of the left end and the
angular acceleration of the rod. Posing these equations in standard
form:
cos q aLEt + H a
= -vLE2/R sin q + w2(L2
- H2)1/2
-sin q aLEt + (L2
- H2)1/2 a = -vLE2/R
cos q - w2
H
Solving these equations using Kramer's rule:
Determinant = cos q (L2
- H2)1/2 + sin q
H
Numerator_1 = { -vLE2/R sin q
+ w2(L2 - H2)1/2
} (L2 - H2)1/2 + { vLE2/R
cos q + w2
H } H
Numerator_2 = -cos q { vLE2/R
cos q + w2
H }+ sin q { -vLE2/R
sin q + w2(L2
- H2)1/2 }
aLEt = Numerator 1 / Determinant = -23.9 m/s2
a = Numerator_2 / Determinant = -36.2
rad/s2
Wow!