1) A homogeneous square plate of mass mp = 20 kg and side length b= 300 mm is attached to a collar of
mass m c = 5 kg by a frictionless pivot at point A. The collar slides without friction on the horizontal
rod. The system is initially at rest when a force P= 100 N is applied to the collar as shown. At this
very instant, compute the angular acceleration a of the square plate.
m c ac
r
y
x
Solving horizontal forces,
Σfx
P
=
=
m(ag)x
mcac + mp(ag)x
100
=
5(ac) + 20(ag)x
1
Distance between the center of gravity and point A,
r
=
lCosθ
Here:
l
θ
=
=
300 mm
450
Thus,
r
=
=
300 Cos450
212.132 mm
Calculating the moment of inertia for the plate about the axis through the center of
gravity,
IG
=
=
=
𝑚𝑝 (2𝑙2 )
12
20(2×3002 )
12
0.3 kgm3
Considering the moment of the point A,
ΣMA
mp(aG) x r
20 x aG x 0.212
4.24 x aG – 0.3 α
=
=
=
=
IGα
0.3 α
0.3 x α
0
2
Applying relative acceleration equation for point A and G,
aG
=
ac + α x rGA – ω2 x rGA
The angular velocity, ω is zero,
Thus,
(aG)xi + (aG)yj =
=
aci + (αk) x (-0.212j) – 0 x (-0.212j)
(ac - 0.212α)i
Equating i coefficients,
(aG)x
ac - (aG)x - 0.212α
=
=
ac - 0.212α
0
3
Equating j coefficients,
(aG)y
=
0
Solving the equations 1 , 2 and 3 ,
ac
aG
α
=
=
=
10 m/s
2.5 m/s
35.33 rad/s2
Therefore, the angular acceleration of the plate is 35.33 rad/s2.
2) The rod of mass m = 30 kg and length l = 1.5 m is released from rest from a
configuration where θ = 45° at which point the spring of stiffness k = 300 N/m is at its
rest length (i.e., it is unstretched). Compute the angular velocity of the rod when θ=0.
A
C
Calculating BC length,
BC
=
1.5 Cos (θ)
At the initial stage,
BC
=
1.5 x Cos (45)
=
1.061 m
=
2π√ 𝑘
=
2π√300
=
1.99 s
=
ut + 2 a t2
Using the equation,
T
Using,
S
𝑚
30
1
Since, the distance travelled by A = distance travelled by B =
=
Using,
1
1.061 =
0 x 1.99 + 2 x a x 1.992
a
=
0.5358 m/s2
V
=
U + at
VA
=
0 + 0.5358 x 1.99
=
1.066 m/s
ω
=
VA x AB
=
1.066 x 1.5
=
1.6 rad/s
Thus,
Rotational Speed,
Therefore, when θ = 0, the rotational speed is 1.6 rad/s.
BC
1.061 m