Rigid Body F=ma Fixed Axis Rotation: Example 3
A 64.4 lb rectangular plate is suspended by a pin and a cord. At the
instant the cord is cut, please determine: (a) Pin reactions at A;
(b) The plate’s angular acceleration, a .
A
4 ft
Cut!
B
G
D
A
3 ft
C
64.4 lb Rectangular plate,
released from rest.
2 ft
2.5 ft
1.5 ft
G
Draw a Free Body Diagram and a Kinetic Diagram
Ay
FBD
KD
4 ft
Ax
G
3 ft =
2.5
ft
3
64.4 lb
m = 64.4/32.2 = 2 slug
4
mat = m
man = m 2 r
=0
Since released
from rest.
r
= 2 (2.5)
=5
Calculate IG for the plate: [IG = (1/12)m(a2 + b2)]
IG = (1/12)(2)(42+32)= 4.17 slug-ft2
Calculate Ipin from the parallel axis theorem:
Ipin = IG + md2 = 4.17 + (2)(2.52) = 16.67 slug-ft2
Ay
Ax
4 ft
0= m
G
3 ft =
2.5
ft
3
64.4 lb
2r
4
mat = m r = 5
Write equations of motion: Sum moments about the pin:
SMpin = Ipina ; (64.4)(2 ft) = Ipina = 16.67a
a = 7.73 rad/sec2
from kinematics:
at = ra = 2.5(7.73), at = 19.3 fps2
Note: Don’t be alarmed about the fact that we drew IGa on
the KD and then used Ipin in the moment equation.
Remember the Ipina term on the RHS is the equivalent of a
kinetic moment of IGa plus mat(r) about the pin. Using Ipin
simplifies the equation slightly.
Ay
Ax
4 ft
0= m
G
3 ft =
2.5
ft
3
64.4 lb
2r
4
mat = 2at = 2 r = 5
Use force summations to find the pin reactions at A:
SFy = may ; Ay – 64.4 = – 2(19.3)(4/5)
Ay = 64.4 – 30.9
Ay = 33.5 lb
SFx = max ; Ax = 2(19.3)(3/5)
Ax = 23.2 lb
As noted earlier, you may choose to express the pin
reactions at A in normal and tangential components,
too. Then sum forces in n and t to compute An and At.
(The mg term would need to be resolved into n and t
coordinates, though.)
Ay
Ax
4 ft
0= m
G
3 ft =
2.5
ft
3
64.4 lb
2r
4
mat = 2at = 2 r = 5
If you choose to sum moments about the pin using a kinetic
moment, (instead of using Ipina) you get:
SMpin = IGa + mat(r); (64.4)(2 ft) = 4.17a + 2at(2.5)
128.8 = 4.17a + 5at ; but here sub in at = 2.5a
128.8 = 4.17a + 5(2.5a) = 16.67a
a = 7.73 rad/sec2
etc.
If you choose to sum moments about G, you introduce the
pin reactions into the moment equation:
SMG = IGa ; Ay(2) – Ax(1.5) = 4.17a
(1)
See the next page for the remainder of this approach….
Ay
4 ft
0= m
Ax
G
3 ft =
2.5
ft
3
64.4 lb
2r
4
mat = 2at = 2 r = 5
Write the F = ma equations:
SFy = may ; Ay – 64.4 = – 2(at)(4/5)
SFx = max ; Ax = 2(at)(3/5)
You need a fourth equation: at = 2.5a.
Solve these
four equations
for the four
unknowns.
2 -1.5 0 -4.17
1 0 1.6 0
0 1 -1.2 0
0 0
1 -2.5
Ay
0
64.4
Ax
=
at
0
0
(2)
(3)
(4)
Ay = 33.5 lb
Ax = 23.2 lb
at = 19.3 fps2
= 7.73 r/s2
These agree with our earlier results, but you can see that
summing moments about G involves more solution effort.
That’s why we recommend summing moments about the pin.