ME625: Assignment 5 solutions

NOTE:

1. The following notation has been used: -

(i) Straight bold small case symbols will be vectors (e.g.: a, b)

(ii) Straight bold capital symbols will be tensors. (e.g. A, B)

2. Make neat Free-Body Diagrams wherever required and solve the questions systemati-
cally with proper equations as taught in class

Problem 1. Find [I]XY Z for the following:

(i) A thin square plate lying in x − y plane. It’s mass is m. It’s edges are 4cm each. A
circular hole of radius r = 0.7cm is cut out as shown.

(ii) A thin wire lies in x − y plane along the curve y = x2 as shown. It’s mass is m.

1
Solution 1. (i) Let’s assume that the width of the plate is b cm and mass m kg. There-
fore, we can write the density of the plate as:
m m m
ρ= 2
= =
16b − πr b 16b − 0.49πb b(16 − πr2 )

We can consider this square plate as combination of a complete square plate with
density ρ and a circular plate with density −ρ as shown in the figure below. This is
also known as principle of superposition.

Now we can calculate the moment of inertia for both the bodies about the origin and
add them to get IXY Z for the given plate. For the first body,
R 2 R R 
(y R + z 2 )dm R − xy dm − R xz dm
[I]cm =  − R xy dm (x2R + z 2 )dm R − yz dm 
− xz dm − yz dm (y 2 + x2 )dm

Since, it is a thin plate, we can put z = 0 in the above formula.

R 2 R 
(y )dm − xy dm 0
[I]1cm = − xy dm
R R
(x2 )dm R 0 
2 2
0 0 (y + x )dm

2
 Figure 1: (i) Figure 2: (ii)

We can consider that the plate is a combination of small rectangular plates of mass
dm. We will calculate [I]11 using the diagram (i) and [I]22 using the diagram (ii) as
shown above.
Z Z 2  3 2
1 2 2 y 64
[I]11 = y dm = y ρ4b dy = 4ρb = ρb
−2 3 −2 3
Z Z 2  3 2
x 64
[I]122 = x2 dm = x2 ρ4b dx = 4ρb = ρb
−2 3 −2 3

Since the plate is symmetric, we know that off-diagonal terms would be 0. Therefore,
 64 
3
ρb 0 0
[I]1cm =  0 64
3
ρb 0 
128
0 0 3
ρb

Now applying parallel axis theorem, we can get [I]XY Z for body 1:

[I]1XY Z = [I]1cm + M 1 [D]1cm/o

 64   2 2

3
ρb 0 0 d 2 + d 3 −d 1 d 2 −d 1 d3
64
= 0 3
ρb 0  + 16ρb  −d1 d2 d21 + d23 −d2 d3 
128
0 0 3
ρb −d1 d3 −d2 d3 d21 + d22
 64   2 2

3
ρb 0 0 3 + 0 −3x3 0
64
= 0 3
ρb 0  + 16ρb  −3x3 32 + 02 0 
128
0 0 3
ρb 0 0 3 + 32
2

3
 144 + 64
 
3
−144 0
[I]1XY Z = ρb  −144 64
144 + 3 0 
128
0 0 288 + 3
 
165.33 −144 0
= ρb  −144 165.33 0 
0 0 330.67

Now, following the same procedure for the disk we can get:

R 2
R 
R(y )dm −
R xy dm 0
[I]2cm = − xy dm (x2 )dm R 0 
2 2
0 0 (y + x )dm

We can consider the disk as a combination of small rings of mass dm. We know that
off-diagonal terms will be zero and [I]211 = [I]222 due to symmetry of disk . We will
calculate [I]11 using the diagram as shown above.
Z r Z 2π
πr4
Z
2 2
[I]11 = y dm = (r sin θ)2 (−ρ)(rdθdrb) = −ρb
0 0 4
Therefore,
 πr4 
4
0 0
πr4
[I]2cm = −ρb  0 4
0 
πr4
0 0 2

4
 Now applying parallel axis theorem, we can get [I]XY Z for body 2:
[I]2XY Z = [I]2cm + M 2 [D]2cm/o
 πr4   2 
4
0 0 d2 + d23 −d1 d2 −d1 d3
4
= −ρb  0 πr4 0  − πr2 ρb  −d1 d2 d21 + d23 −d2 d3 
0 0 πr2
4
−d1 d3 −d2 d3 d21 + d22
 r2   2 2

4
0 0 3 + 0 −(3 + r)x3 0
2
= −πr2 ρb  0 r4 0  − πr2 ρb −(3 + r)x3 (3 + r)2 + 02 0 
r2 2 2
0 0 2 0 0 (3 + r) + 3
 
9.1225 −11.1 0
= −0.49πρb −11.1 13.8125
 0 
0 0 22.935
Now applying principle of superposition, we can write:
[I]XY Z = [I]1XY Z + [I]2XY Z
   
165.33 −144 0 9.1225 −11.1 0
= ρb  −144 165.33 0  − 0.49πρb  −11.1 13.8125 0 
0 0 330.67 0 0 22.935
   
151.29 −126.91 0 151.29 −126.91 0
m 
[I]XY Z = ρb −126.91 144.07 0 = −126.91 144.07 0 
14.46
0 0 295.36 0 0 295.36
 
10.46 −8.78 0
[I]XY Z = m −8.78 9.96 0  kgcm2
0 0 20.43

(ii) Since we need to know the linear density(λ) of the wire, we will start by finding it’s
length:
dy d 2
= (x ) = 2x = tan θ
dx Z dx Z Z 0.5 √
0.5
dx
l = dl = = 1 + 4x2 dx = 0.574 m
0 cos θ 0
m m
λ= = kg/m
l 0.574
Now, we can calculate moment of inertia:
R 2 R R 
(y R + z 2 )dm R − xy dm − R xz dm
[I]XY Z =  − R xy dm (x2R + z 2 )dm R − yz dm 
− xz dm − yz dm (y 2 + x2 )dm
Since, it is a planar problem, we can put z = 0 in the above formula.
R 2 R 
R(y )dm − R xy dm 0
[I]XY Z = − xy dm (x2 )dm R 0 
2 2
0 0 (y + x )dm

5
 We can consider the wire to be a combination of point masses of mass dm = λdl. We
will calculate the components using the diagram shown above:
Z
2
Z 0.574
2
Z 0.5 √
[I]11 = y dm = y λdl = x4 λ 1 + 4x2 dx = 0.0082λ = 0.0143m
Z Z0 0.574 Z0 0.5 √
[I]22 = x2 dm = x2 λdl = x2 λ 1 + 4x2 dx = 0.0525λ = 0.0915m
0 0
Z Z 0.574 Z 0.5 √
[I]12 = − xy dm = − xyλdl = − x3 λ 1 + 4x2 dx = −0.0201λ = −0.0350m
0 0

 
0.0143 −0.0350 0
[I]XY Z = m −0.0350 0.0915 0  kgm2
0 0 0.1058

Problem 2. A uniform rod of mass m and length l attached to a uniform disk of mass
M and radius R at point P (centre of disk), is released from rest. The hinge at P can be
considered frictionless. Immediately after the release, find the normal reaction force from
the wall at Q. The schematic diagram can be seen below.
Hint: Apply LMB and then AMB about point C.

6
Solution 2. Let’s start the problem by noting what we know about the problem,

(i) If acceleration of Point P is ap then ap · êj = 0 =⇒ ap = −ap êi

(ii) If acceleration of Point Q is aq then aq · êi = 0 =⇒ aq = −aq êj

(iii) If angular acceleration of disk is αdisk and there is no slip condition at Point C then
aC = 0 =⇒ ap + αdisk × P C = 0 =⇒ ap = −αdisk êk × −Rêj = −αdisk Rêi . So we can
write, ap = αdisk R

(iv) Since the system is released from rest, the angular velocities of the rod and the disk
are 0 =⇒ wrod = wdisk = 0.

7
Now in disk, we have point P as centre of mass of disk and we can apply AMB about point
C:
X X
ri/C × F i + M j = rP/C × M aP + Icm · αdisk + wdisk × Icm · wdisk
i j

We know that,
1

2 0 0
MR 2 1 
Icm = 0 2 0 and wdisk = 0
2
0 0 1

We will only have horizontal component of Reaction force at P in the AMB equation

CP × H P + 0 = CP × M aP + Icm · αdisk + 0
M R2
Rêj × (−HP x êi − HP y êj ) = Rêj × M (−aP )êi + (êi ⊗ êi + êj ⊗ êj + 2êk ⊗ êk ) · αdisk êk
4
M R2 M R2
RHP x êk = M RaP êk + αdisk êk = M R(αdisk R)êk + αdisk êk
2 2
3
=⇒ HP x = M Rαdisk (1)
2
Now moving to the FBD of the rod, we can consider the acceleration of the COM of the rod
as some arbitrary vector aG and we can write the accelerations of Point P and Point Q on
rod in terms of aG :

aP = aG + αrod × rP/G ∵ wrod = 0

L L
= aG + (−αrod )êk × (− cos θêi − sin θêj )
2 2
αrod L αrod L
= aG + cos θêj − sin θêi
2 2
Similarly for point Q,

aQ = aG + αrod × rQ/G ∵ wrod = 0

L L
= aG + (−αrod )êk × ( cos θêi + sin θêj )
2 2
αrod L αrod L
= aG − cos θêj + sin θêi
2 2
αrod L αrod L αrod L αrod L
= aP − cos θêj + sin θêi − cos θêj + sin θêi
2 2 2 2
−aQ êj = aP − αrod L cos θêj + αrod L sin θêi = −aP êi + αrod L sin θêi − αrod L cos θêj
=⇒ aQ = αrod L cos θ and aP = αrod L sin θ (2)
αrod L αrod L
=⇒ aG = − sin θêi − cos θêj (3)
2 2

8
Now, we will apply LMB on rod in êi direction
X
F i · êi = maG · êi
i
αrod L maP mRαdisk
HP x − N = −m sin θ = − =−
2 2 2
(3M + m) (3M + m) (3M + m)
N= Rαdisk =⇒ N = − ap êi = − αrod L sin θêi (4)
2 2 2
Applying AMB about Point P on rod:
X X
ri/P × F i + M j = rG/P × maG + Icm · αrod + wrod × Icm · wrod
i j

rG/P × (−mg)êj + rQ/P × (−N )êi = rG/P × maG + Icm · αrod

sin2 θ
 
2 − sin θ cos θ 0
mL 
Icm = − sin θ cos θ cos2 θ 0
12
0 0 1
2
  
sin θ − sin θ cos θ 0 0
mL2 
Icm · αrod = − sin θ cos θ cos2 θ 0  0 
12
0 0 1 −αrod
mL2
=− αrod êk
12

 
L L
cos θêi + sin θêj × (−mg)êj + (L cos θêi + L sin θêj ) × (−N )êi =
2 2
mL2
   
L L αrod L αrod L
cos θêi + sin θêj × m − sin θêi − cos θêj − αrod êk
2 2 2 2 12
mgL cos θ mL2 αrod cos2 θ mL2 αrod sin2 θ mL2
− êk + N L sin θêk = − êk + êk − αrod êk
2 4 4 12
mgL cos θ mL2
− êk + N L sin θêk = − αrod (1 + 3 cos 2θ) êk
2 12

9
Now using eq. (4) in the above expression

mgL cos θ mL2 2N

− + N L sin θ = − (1 + 3 cos 2θ)
2 12 (3M + m)L sin θ
mgL cos θ 1
N= mL(1+3 cos 2θ)
2 + L sin θ
6(3M +m) sin θ
mg cos θ 6(3M + m) sin θ
=
2 m(1 + 3 cos 2θ) + 6(3M + m) sin2 θ
3 (3M + m)mg sin 2θ
=
2 m(1 + 3 cos 2θ) + 3(3M + m)(1 − cos 2θ)
3 (3M + m)mg sin 2θ
=
2 [m + 3m cos 2θ + 9M (1 − cos 2θ) + 3m(1 − cos 2θ)]
3 (3M + m)mg sin 2θ
N=
2 [4m + 9M (1 − cos 2θ)]

3 (3M + m)mg sin 2θ

N =− êi
2 [4m + 9M (1 − cos 2θ)]

Problem 3. The bar CD in the figure below is rotated at a constant angular velocity ω 0 ,
which causes the collar B to slide along its length and in turn rotate the bar AB attached
to the collar.
For the instant shown, find the following kinematic quantities:
(i) Angular velocity ω of the rod AB.

(ii) Velocity of collar B.

(iii) Acceleration of collar B.

Solution 3. We start by assigning frames to different links. We define our base frame
{Eo , A, Êi } and the rotating frames rigidly attached to links AB and CD as {A, A, âi },
{C, C, ĉi } respectively.

10
 π
Let the relative angle between Eo and A be 2
+ θ and that between Eo and C be β. Then
the frame transformations can be written as:
⇒ â1 = − sin(θ)Ê1 + cos(θ)Ê2
â2 = − cos(θ)Ê1 − sin(θ)Ê2
⇒ ĉ1 = cos(β)Ê1 + sin(β)Ê2
ĉ2 = − sin(β)Ê1 + cos(β)Ê2

The angular velocities of the frames as seen from ground frame are:
ωA/Eo = −ωâ3 = −ω Ê3
ωC/Eo = ω 0 ĉ3 = ω 0 Ê3
Now, the velocity of point B as seen from A is:
vB/A = ωA/Eo × rB/A
= −ωâ3 × lâ1
= −ωlâ2
= ωl cos(θ)Ê1 + ωl sin(θ)Ê2
Velocity of point P on bar CD that coincides with point B at the given moment is given as:
vP/C = vrel + ωC/Eo × rB/C
= vr Ê1 + ω 0 ĉ3 × (Rĉ1 + Rĉ2 )
= vr Ê1 + ω 0 Rĉ2 − ω 0 Rĉ1
= vr Ê1 + ω 0 R[− sin(β)Ê1 + cos(β)Ê2 ] − ω 0 R[cos(β) + sin(β)Ê2 ]
= [vr − ω 0 R(sin(β) + cos(β))]Ê1 + ω 0 R[cos(β) − sin(β)]Ê2

11
At the given moment, the vertical velocities (in the Ê2 of points B and P will be the same
(due to the rod constraining their motion). Therefore:
ωl sin(θ) = ω 0 R[cos(β) − sin(β)]
 
4 0 3R
ω= ω ∵ β = 0 and sin(θ) =
3 4l
Using this relation, the angular velocity of point B can be found using the equation for
vB/A :
 
4 0 R 3R
vB/A = ω l Ê1 + Ê2
3 l 4l
4
vB/A = ω 0 R[ Ê1 + Ê2 ]
3
Now using this and comparing the velocity components in the Ê1 direction, we get vrel :
7
vrel = ω 0 RÊ1
3
Now, let the angular acceleration of bar AB be αâ3 . Then the acceleration of point B as
seen from A can be written as:
aB/A = αA × rB/A + ωA/Eo × (ωA/Eo × rB/A )
= αlâ2 − ω 2 lâ1
= [−αl cos(θ) + ω 2 l sin(θ)]Ê1 + [−αl sin(θ) − ω 2 l cos(θ)]Ê2
and the acceleration of point P as seen from C,
aP/C = arel + αC/Eo × rP/C + ωC/Eo × (ωC/Eo × rP/C ) + 2ωC/Eo × vrel
14
= ar Ê1 + ω 0 ĉ3 × [ω 0 R(ĉ2 − ĉ1 )] + ω 02 Rĉ2
3
14
= [ar − ω 02 R(cos(β) − sin(β)) − ω 02 R sin(β)]Ê1
3
14
+ [−ω 02 R(sin(β) + cos(β)) + ω 02 R cos(β)]Ê2
3
Again, due to the constraint between the two links, the vertical acceleration of point B
and P will be the same:
14
−αl sin(θ) − ω 2 l cos(θ) = −ω 02 R(sin(β) + cos(β)) + ω 02 R
3
02
ω R(sin(β) + cos(β)) − ω l cos(θ) − 14
2
3
ω 02 R
α=
l sin(θ)
196 02
=− ω
27
put this term in the acceleration equation for B and replacing the values of θ and β in
it, we get the final acceleration as:
 
02 232 11
aB/A = ω R Ê1 + Ê2
27 3

12
Problem 4. The uniform slender bar of mass m and length l is centrally mounted on the
shaft A-A, about Which it rotates with a constant speed φ̇ = p. Simultaneously, the yoke is
forced to rotate about the x-axis with a constant speed ω0 . As a function of φ, determine
the magnitude of the torque M required to maintain the constant speed ω0 .

Solution 4. Let us consider three frames at O, ground frame {E0 , O, Êi }, frame attached
to yoke {E, O, êi } and BFCS {E0 , O, eˆ0 i }. Hence

ω E0 /E0 ]mix = φ̇eˆ0 3 + ω0 Ê1

[ω
ω E0 /E0 ]E0 = φ̇Ê3 + ω0 Ê1
[ω
also,
d d
αE0 /E0 ]mix =
[α ω E0 /E0 × eˆ03 )
(φ̇eˆ0 3 ) = φ̇ (eˆ0 3 ) = φ̇(ω
dt dt
αE0 /E0 ]E0 = φ̇(ω0 Ê1 × Ê3 ) = −φ̇ω0 Ê2
[α

13
For AMB about O, as vO = 0, aO = 0, we can write
X X
[ri/P ]E0 × [Fi ]E0 + [Mj ]E0 = [Icm ]E0 · [α ω ]E0 × [Icm ]E0 · [ω
α]E0 + [ω ω ]E0
i j

To compute [Icm ]E0 :

We know in BFCS  
20 0 0
ml 
[Icm ]E0 = 0 1 0
12
0 0 1
[Icm ]E0 = [A]T · [Icm ] · [A]
   
2 cos(φ) −sin(φ) 0 0 0 0 cos(φ) sin(φ) 0
ml 
[Icm ]E0 = sin(φ) cos(φ) 0 0 1 0 −sin(φ) cos(φ) 0
12
0 0 1 0 0 1 0 0 1
 
sin2 (φ) −sin(φ)cos(φ) 0
ml2 
[Icm ]E0 = −sin(φ)cos(φ) cos2 (φ) 0
12
0 0 1
substituting in AMB, we get
    
Mx 2 sin2 (φ) −sin(φ)cos(φ) 0 0
My  = ml −sin(φ)cos(φ) cos2 (φ) 0 −φ̇ω0  + [ω ω ]E0 × [Icm ]E0 · [ω
ω ]E0
12
Mz 0 0 1 0
 
φ̇ω sin (φ) cos (φ)
ml2  o ml2
= −φ̇ωo cos2 (φ)  + (ω0 Ê1 + φ̇Ê3 ) × (sin2 (φ) ωo Ê1 − sin (φ) cos (φ) ωo φ̇Ê2 + φ̇Ê3 )
12 12
0
   
2 φ̇ωo sin (φ) cos (φ) 2 φ̇ωo sin (φ) cos (φ)
ml  ml 
= −φ̇ωo cos2 (φ)  + φ̇ωo cos2 (φ) 
12 12
0 −φ̇ωo sin (φ) cos (φ)

On comparing the first elements in LHS and RHS, we get

ml2 ml2
Mx = M = φ̇ω0 sin(φ)cos(φ) = φ̇ω0 sin(2φ)
6 12
Problem 5. The solid circular disk of mass m and small thickness is spinning freely on
its shaft at the rate p. If the assembly is released in the vertical position at θ = 0 with
θ̇ = 0, determine the horizontal components of the forces A and B exerted by the respective
bearings on the horizontal shaft as the position θ = π/2 is passed. Neglect the mass of the
two shafts compared with m and neglect all friction. Solve by using the appropriate moment
equations. [15 marks]

14
Solution 5. Let us attach frames to the given system as shown:

Here, {Eo , O, Êi } is the base frame, {E0 , O, ê0i } is the rotating frame attached to the shaft
OG and {E, G, êi } is the frame rigidly attached to the disk.
We start by calculating the kinematic quantities for point G. The angular velocities of

15
the rotating frames can be written as:

ωE0 /Eo = θ̇Ê2 = θ̇ê02

ωE/Eo = θ̇Ê2 + pê03 = θ̇ê02 + pê03

The angular acceleration then becomes:

δ θ̇ê02 δpê03
αE/Eo = + ωE0 /Eo × θ̇ê02 + + ωE0 /Eo × pê03
δt δt
= θ̈ê02 + ṗê03 + (θ̇ê02 × pê03 )
= θ̈ê02 + ṗê03 + θ̇pê01

The position of point G is rG/O = lê03 . Using this, we can write the velocity of this point
as:

vG/O = ωE/Eo × rG/O

vG/O = (θ̇Ê2 + pê03 ) × lê03
vG/O = θ̇lê01

and the acceleration as:

aG/O = aO + arel + αE/Eo × rG/O + ωE/Eo × (ωE/Eo × rG/O ) + 2vrel × αE/Eo

= (θ̈ê02 + ṗê03 + θ̇pê01 ) × lê03 + ωE/Eo × (ωE/Eo × rG/O )
= −θ̇plê02 + θ̈lê01 + θ̇plê02 − θ̇2 lê03
= θ̈lê01 − θ̇2 lê03

Now, we will apply the momentum balance laws on the system. Let the reaction forces
at points A and B be:
RA = RAx Ê1 + RAz Ê3
RB = RBx Ê1 + RBz Ê3
Now, Applying LMB gives us:
X
F = maG/O
RA + RB − mg Ê3 = maG/O
= m[θ̈lê01 − θ̇2 lê03 ]
˙ 2 l cos(θ))Ê3 ]
(RAx + RBx )Ê1 + (RAz + RBz − mg)Ê3 = m[(θ̈l cos(θ) − θ̇2 l sin(θ))Ê1 + (−θ̈l sin(θ) − (θ)

comparing the coefficients both sides:

RAx + RBx = m[θ̈l cos(θ) − θ̇2 l sin(θ)]

RAz + RBz = m[−θ̈l sin(θ) − θ̇2 l cos(θ)] + mg

16
 Applying AMB and writing components in E0 :
X
MO = rG/O × maG/O + Icm · αE/Eo + ωE/Eo × Icm · ωE/Eo

LHS := RA × (−bÊ2 ) + RB × (bÊ2 ) + (−mg Ê3 ) × lê03

= (−RA + RB ) × bÊ2 − mglÊ3 × ê03
= −(RBz − RAz )bÊ1 − mgl sin(θ)Ê2 + (RBx − RAx )bÊ3
RHS := lê03 × m(θ̈lê01 − θ̇2 lê03 ) + [Icm ]E0 [αE/Eo ]E0 + [ωE/Eo ]E0 × [Icm ]E0 [ωE/Eo ]E0
     
0 −1 0 1 2 1 0 0 θ̇p
mr 
= mθ̈l2 1 0 0 0 + 0 1 0  θ̈ 
4
0 0 0 0 0 0 2 ṗ
   
0 −p θ̇ 1 0 0 0
mr2 
+ p 0 0   0 1 0   θ̇ 
4
−θ̇ 0 0 0 0 2 p
mr2 mr2 0 mr2 0 mr2
= mθ̈l2 ê02 + θ̇pê01 + θ̈ê2 + ṗê3 + [−pθ̇ + 2pθ̇]ê01
4 4 4 4
mr2 mr2 0 mr2
= (mθ̈l2 + θ̈)ê02 + ṗê3 + θ̇pê01
4 2 2
mr2 mr2 mr2
= (mθ̈l2 + θ̈)ê02 + ṗ[sin(θ)Ê1 + cos(θ)Ê3 ] + θ̇p[cos(θ)Ê1 − sin(θ)Ê3 ]
4 2 2
mr2 mr2 mr2
 
= ṗ[ṗ sin(θ) + θ̇p cos(θ)]Ê1 + mθ̈l2 + θ̈ Ê2 + [ṗ cos(θ) − θ̇p sin(θ)]Ê3
2 4 2

comparing the components corresponding to the respective basis in LHS and RHS above,
we get the following set of equations:

mr2
RBx − RAx = [ṗ cos(θ) − θ̇p sin(θ)]
2b
mr2
mθ̈l2 + θ̈ + mgl sin(θ) = 0
4
mr2
RBx − RAx = − [ṗ cos(θ) − θ̇p sin(θ)]
2b
Rewriting the second equation above, we get:

r2
 
2
l + θ̈ + gl sin(θ) = 0
4
" #
gl
θ̈ = −ω 2 sin(θ) ω2 = 2
l + r4
2

dθ̇θ̇ = −ω 2 sin(θ)dθ
p
θ̇ = ω 2(1 − cos(θ))

17
at θ = π2 ,
√
θ̇ = ω 2
θ̈ = −ω 2

Now, adding the first component equations of LMB and AMB and putting in the values
of θ, θ̇ and θ̈, we get:
" # " s #
mg mr2 p gl
RBx = − 2 − √ 2
1 + 4lr 2 2 2b l2 + r4
" # " s #
mg mr2 p gl
RAx = − 2 + √ 2
1 + 4lr 2 2 2b l2 + r4

Notice the extra reaction terms at A and B. These are due to the extra moment developed
due to non-zero angular momentum of the disc that gives rise to gyroscopic precession.

Problem 6. A uniform slender rod AB of mass m and length L is welded at A to a rotating

disc, spinning about its axis at a constant angular velocity ω , as shown. Determine the value
of ω which will result in zero moment at the weld at A. The gravity acts vertically downward.

Solution 6. As the point B coinciding with the central axis of the body, hence vB = 0, let
us apply the LMB, and the AMB about this point.

FBD:-

18
LMB:-
X
F = maG


= m aB + arel + wE/Eo × (wE/Eo × rA/C ) + αE/Eo × rA/C + 2wE/Eo × vrel
0  0
 
=m  aB
*+ a *
 + w × (w × r ) + α :
 0
× r + 2w × v :
 0
rel E/Eo E/Eo A/C  A/C rel
E/Eo
 E/Eo 


= m wE/Eo × (wE/Eo × rA/C )

In the frame {E0 , B, Êi }, we can write

X  
[F]E0 = m [wE/Eo ]E0 × ([wE/Eo ]E0 × [rA/C ]E0 )
  
L
R1 Ê1 + R2 Ê2 + R3 Ê3 − mg Ê2 = m ω Ê2 × ω Ê2 × (− cos θÊ1 + sin θÊ2 )
2
2
mω L
R1 Ê1 + (R2 − mg)Ê2 + R3 Ê3 = cos θÊ1
2
On comparing LHS and RHS, we get
l
R1 = mω 2 cos θ R2 = mg and R3 = 0
2

19
AMB:-
X X
ri/B × Fi + Mi = rG/B × maB + IB · α E/Eo + wE/Eo × IB · wE/Eo
0 0w
= rG/B × maB
*+ IB · 
α E/E
 :+

o E/Eo × IB · wE/Eo
= wE/Eo × IB · wE/Eo

LHS of AMB, in the frame {E0 , B, Êi } :-

X X
L.H.S. = [ri/B ]E0 × [Fi ]E0 + [Mi ]E0
mgL cos θ 0 0 0
= −R1 L sin θÊ3 − R2 L cos θÊ3 + Ê3 + 
M>Ê + M
1 1
>Ê + M
2 2 3 3
>Ê
2
2 2
 
mω L mgL cos θ
= − cos θ sin θ − mgL cos θ + Ê3 (computing R1 , R2 and R3 from above)
2 2

RHS of AMB, in the frame {E0 , B, Êi } :-

R.H.S. = [wE/Eo ]E0 × [IB ]E0 · [wE/Eo ]E0

 
= [wE/Eo ]E0 × [I12 ]B ω Ê1 + [I22 ]B ω Ê2 + [I32 ]B ω Ê3
 
= ω Ê2 × [I12 ]B ω Ê1 + [I22 ]B ω Ê2 + [I32 ]B ω Ê3
= ω 2 ([I32 ]B Ê1 − [I12 ]B Ê3 )

Now,

20
 Z L Z L
I32 = − x3 x2 dm = − 0 · x2 dm = 0
0 0
and
L L
mL2
Z Z
m
I12 = − x1 x2 dm = − (−p cos θ)(p sin θ) dp = cos θ sin θ
0 0 L 3
Substituting in R.H.S., we get
2 mω 2 L2
R.H.S. = −ω I12 Ê3 = − cos θ sin θÊ3
3
finally, on equating L.H.S. and R.H.S., we get
mω 2 L2 mω 2 L2
 
mgL cos θ
− cos θ sin θ − mgL cos θ + Ê3 = − cos θ sin θÊ3
2 2 3
r
3g
ω= − (imaginary value)
L sin θ
NOTE:
(i) If we could not have made M = 0, in that case, we get
mω 2 L2 mω 2 L2
 
mgL cos θ
M3 − cos θ sin θ − mgL cos θ + Ê3 = − cos θ sin θÊ3
2 2 3
mgL mω 2 L2
M3 − cos θ = cos θ sin θ
2 6
So, for ω = 0, we get static moment reaction
mgL
M3 =cos θ
2
and for any ω > 0, the moment reaction will increase, so if we make M or M3 = 0, ω should
be imaginary indeed.
(ii) If we take θ in negative direction, then ω is real, you can check for this orientation.

21
Problem 7. A 50 kg cylindrical rotor is mounted on two bearings A and B of a massless
bracket as shown. The bracket is connected to a motor which spins it at a constant rate of
Ω = 10rad/sec as shown. The rotor is made to spin at a constant rate p = 50rad/sec in the
direction shown. For a = 100 mm and r = 50 mm determine the following:

1. Bending moment in the shaft at C expressed in E0 .

2. Kinetic energy of the rotor.

For your convenience the BFCS E to be used in calculations is shown in the figure.

Solution 7. The given frames at the instant are related as:

 
1 0 0
[êi ] = 0 cos(π − π3 ) − sin(π − π3 ) [Êi ]
0 sin(π − π3 ) cos(π − π3 )
 
1 0 0
= 0 − cos( π3 ) − sin( π3 )  [Êi ]
0 sin( π3 ) − cos( π3 )
⇒ ê1 = Ê1
√
1 3
ê2 = − Ê2 + Ê3
2
√ 2
3 1
ê3 = − Ê2 − Ê3
2 2

22
The angular velocity of frame E is:

ωE/Eo = ΩÊ3 + pê3

√ !
3 1
=Ω ê2 − ê3 + pê3
2 2
√  
3 Ω
= Ωê2 + p − ê3
2 2
= 8.66ê2 + 45ê3

and angular acceleration:

αE/Eo = ωE/Eo × pê3

= ΩÊ3 × pê3
√ !
3 1
= Ωp ê2 − ê3 × ê3
2 2
= 433.01ê1

Now in the angular moment balance equation:

X
M = I · α + ω × (I · ω)

The net external moment in the LHS is entirely due to the reaction forces at A and B in
this problem. Therefore, this is the bending moment that we have to calculate.

M = [I]E [αE/Eo ]E + [ωE/Eo ]E × [I]E [·ωE/Eo ]E (5)

For the given cylinder, the moment of inertia is:

 M a2 M r2 
3
+ 4 0 0
M a2 2
[I]E =  0 3
+ M4r 0 
M r2
0 0
  2
0.1979 0 0
= 0 0.1979 0 
0 0 0.0625

23
Putting this and the angular velocity and acceleration terms above in equation (1), we get:
  
0.1979 0 0 433.01
[M]E =  0 0.1979 0  0 
0 0 0.0625 0
   
0 −45 8.66 0.1979 0 0 0
+  45 0 0  0 0.1979 0  8.66
−8.66 0 0 0 0 0.0625 45
 
32.93
= 0 
0
= 32.93ê1
= 32.93Ê1

Net bending moment is therefore: 32.93 N-m.

For the second part, the kinetic energy can be calculated as:
1 T
T = ωE/E o
IωE/Eo
2
1
= [ωE/Eo ]TE [IE ][ωE/Eo ]E
2   
0.1979 0 0 0
1 
= 0 8.66 45  0 0.1979 0  8.66
2
0 0 0.0625 45
= 70.70 J

Problem 8. The half-cylindrical shell of radius r, length 2b, and mass m revolves about the
vertical z -axis with a constant angular velocity ω as indicated. Determine the magnitude M
of the bending moment in the shaft at A due to both the weight and the rotational motion
of the shell.

24
Solution 8. As we have to find bending moment in shaft at A, we will consider a frame
{E0 , A, Êi } at this point.

Let A1 , A2 and A3 be the reaction forces at point A. We don’t need to find these values
as reaction moment can be evaluated directly by considering AMB about point A.
AMB:-
X X
ri/A × Fi + Mi = rG/A × maA + IA · α E/Eo + wE/Eo × IA · wE/Eo

= rG/A × m 0 I ·α  0w
aA
*+ E/Eo × IA · wE/Eo

A E/E
:+
o

= wE/Eo × IA · wE/Eo

LHS of AMB, in the frame {E0 , A, Êi } :-

X X
L.H.S. = [ri/A ]E0 × [Fi ]E0 + [Mi ]E0
2mgr
= M1 Ê1 + M2 Ê2 + M3 Ê3 + Ê1
π

RHS of AMB, in the frame {E0 , B, Êi } :-

R.H.S. = [wE/Eo ]E0 × [IA ]E0 · [wE/Eo ]E0

 
= [wE/Eo ]E0 × [I13 ]A ω Ê1 + [I23 ]A ω Ê2 + [I33 ]A ω Ê3
 
= ω Ê3 × [I13 ]A ω Ê1 + [I23 ]A ω Ê2 + [I33 ]A ω Ê3
= ω 2 (−[I23 ]A Ê1 + [I13 ]A Ê2 )

25
 Now, from the above figure
Z
[I13 ]A = − x1 x3 dm
Zs
=− x1 (2r + r cos θ)ρ dA
Zs
=− x1 (2r + r cos θ)ρ rdθ dx1
s
Z πZ b
2
= −ρr x1 (2 + cos θ) dθ dx1
0 −b
π x1 =b
x2
Z 
= −ρr 2
(2 + cos θ) dθ · 1 =0
0 2 x1 =−b

and,
Z
[I23 ]A = − x2 x3 dm
Zs
=− −r sin θ(2r + r cos θ)ρ dA
Zs
=− −r sin θ(2r + r cos θ)ρ r dθ dx1
s
Z πZ b
3
= ρr sin θ(2 + cos θ) dθ dx1
0 −b
Z π
3
= 2bρr sin θ(2 + cos θ) dθ dx1
0
4mr2 m
= 8bρr3 = (as, ρ = )
π 2πrb

26
Now, substituting all these values in LHS and RHS of AMB, we get:
2mgr
M1 Ê1 + M2 Ê2 + M3 Ê3 + Ê1 = ω 2 (−[I23 ]A Ê1 + [I13 ]A Ê2 )
π
2mgr *0
Ê1 = ω 2 (−[I23 ]A Ê1 + 

M1 Ê1 + M2 Ê2 + M3 Ê3 + [I13
] A Ê2 )
π
2mgr 4mω 2 r2
M1 Ê1 + M2 Ê2 + M3 Ê3 = − Ê1 − Ê1
π π
On comparing, we get

2mgr 4mω 2 r2
 
2mr
M1 = − + =− (g + 2ω 2 r)
π π π
M2 = 0
M3 = 0

Hence the required moment magnitude is:

2mr
|M| = (g + 2ω 2 r)
π

27