Problem 5/63 Page 360, Engineering Mechanics - Dynamics, Meriam and Kraige, 4th Edition
Given: A bar of length L with right end moving with a velocity of V0 to the right and left end moving along a circle of radius R. At the instant of interest the left end 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 of q with the horizontal.
Find: The angular speed of the bar and the speed 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 velocities of the two ends of the bar are related by the velocity
two point formula. Considering +X to be horizontal to the right
and +Y to be vertically upward:
vRE = vLE + w X r
V0 i = vLE ( cos q i
- sin q j
) + w k
X ( (L2 - H2)1/2 i - H j
)
V0 i = vLE ( cos
q i
- sin q
j ) + w(L2
- H2 )1/2 j + w
H i
Equating components:
V0 = vLE cos q
+ wH
0 = - vLE sin q
+ w(L2
- H2 )1/2
Solving the second equation for and plugging into the first equation:
V0 = vLE cos q
+ vLE sin q H
/ (L2 - H2 )1/2
Solving for the speed of the left end:
vLE = V0 / { cos q
+ sin q H
/ (L2 - H2 )1/2 }
= 4.38 m/s
Plugging this result into the second
equation and solving for the angular speed:
w = [ sin q
V0 / { cos q
+ sin q H
/ (L2 - H2 )1/2 }]
/ (L2 - H2 )1/2 = 3.23 rad/s