β
r = 3 in A
ω ω
Problem 18.154
A thin ring of 3-in. radius is attached by a collar at A to a vertical shaft which rotates at the constant rate ω.
Determine (a) the constant angle β that the plane of the ring forms with the vertical when ω = 12 rad/s, (b) the maximum value of ω for which the ring will remain vertical (β = 0 ).
Solving Problems on Your Own Problem 18.154
β
r = 3 in A
ω
ω A thin ring of 3-in. radius is attached by a collar at A to a vertical shaft which rotates at the constant rate ω. Determine (a) the constant angle β that the plane of the ring forms with the vertical when ω = 12 rad/s, (b) the maximum value of ω for which the ring will remain vertical (β = 0 ).
1. Determine the angular velocity ωω of the body and the angular velocity ΩΩ of the rotating frame: ω is the angular velocity of theω body with respect to a fixed frame of reference. The vector ωω may be resolved into components along the rotating axes. ΩΩ is the angular velocity of the rotating frame. If the rotating frame is rigidly attached to the body, ΩΩ = ω ω .
Solving Problems on Your Own Problem 18.154
β
r = 3 in A
ω
ω A thin ring of 3-in. radius is attached by a collar at A to a vertical shaft which rotates at the constant rate ω. Determine (a) the constant angle β that the plane of the ring forms with the vertical when ω = 12 rad/s, (b) the maximum value of ω for which the ring will remain vertical (β = 0 ).
2. Determine the mass moments and products of inertia of the body: For a three dimensional body these are the quantities Ix, Iy, Iz, Ixy, Ixz, and Iyz, where xyz is the rotating frame. If the rotating frame is centered at G (mass center) and is in the direction of the principal axes of inertia (Gx’y’z’), then the products of inertia are zero and Ix, Iy, and Iz, are the principal centroidal moments of inertia.
Solving Problems on Your Own Problem 18.154
β
r = 3 in A
ω
ω A thin ring of 3-in. radius is attached by a collar at A to a vertical shaft which rotates at the constant rate ω. Determine (a) the constant angle β that the plane of the ring forms with the vertical when ω = 12 rad/s, (b) the maximum value of ω for which the ring will remain vertical (β = 0 ).
3. Determine the angular momentum of the body: The angular momentum HGof a rigid body about the principal axes of inertia (Gx’y’z’) is given terms of the components of its angular velocity ω
ω and its principal centroidal moments of inertia.
Hx’= Ix’ωx’ Hy’= Iy’ωy’ Hz’= Iz’ωz’
Solving Problems on Your Own Problem 18.154
β
r = 3 in A
ω
ω A thin ring of 3-in. radius is attached by a collar at A to a vertical shaft which rotates at the constant rate ω. Determine (a) the constant angle β that the plane of the ring forms with the vertical when ω = 12 rad/s, (b) the maximum value of ω for which the ring will remain vertical (β = 0 ).
4. Compute the rate of change of angular momentum : The rate of change of HGwith respect to a fixed frame is given by
HG= ( HG)Oxyz+ ΩΩ x HG
where ( HG)Oxyzis the rate of change of HGwith respect to the rotating frame, and ΩΩ is the angular velocity of the rotating frame.
If the rotating frame is rigidly attached to the body, ΩΩ is equal to ωω, the angular velocity of the body.
. .
Solving Problems on Your Own Problem 18.154
β
r = 3 in A
ω
ω A thin ring of 3-in. radius is attached by a collar at A to a vertical shaft which rotates at the constant rate ω. Determine (a) the constant angle β that the plane of the ring forms with the vertical when ω = 12 rad/s, (b) the maximum value of ω for which the ring will remain vertical (β = 0 ).
5. Draw the free-body-diagram equation: The diagram shows that the system of the external forces exerted on the body is equivalent to the vector ma applied at G and the couple vector HG.
6. Write equations of motion: Six independent scalar equations can be written from
ΣF = m a, ΣMG= H. G
.
Problem 18.154 Solution Determine the angular velocity ωω of the body.
β
r = 3 in A
ω ω
β
r = 3 in x’
y’
z’
G
Use the principal axes Gx’y’z’, with x’ to the plane of the ring.
ω
ω ωx’= ω sin β
ωy’= ω cos β ωz’= 0
Determine the angular velocity ΩΩ of the rotating frame.
Determine the mass moments of inertia.
ω
ω = ω sinβ i’ + ω cosβ j’
Ω Ω = ωω
Ix’= m r2 Iy’= m r2 Iz’ = m r2
1 12 2
Problem 18.154 Solution
β
r = 3 in A
ω ω
β
r = 3 in x’
y’
z’
G ω ω
Determine the angular momentum of the body.
HG= Ix’ ωx’i’ + Iy’ωy’j’ + Iz’ωz’k’
HG= m r2ω ( sin β i’ + cosβ j’ )12 Compute the rate of change of angular momentum.
Ω
Ω = ω sinβ i’ + ω cosβ j’
Ω
Ω x HG= (mr2ω ) i’ j’ k’
ω sinβ ω cosβ 0 sin β 12cosβ 0 H
.
G= - 1 mr2ω sinβ cosβ k’2
H
.
G= ( H.
G)oxyz+ ΩΩ x HG( H
.
G)oxyz = 0, since ωω = constantProblem 18.154 Solution
Draw the free-body-diagram equation.
β
r = 3 in A
ω ω
A
β
r sinβ W
Ax Ay
A
G
m a = m ω2 (r sinβ ) H
.
Gr cosβ
=
Problem 18.154 Solution A
β
r sinβ W
Ax Ay
A
G
m a = m ω2 (r sinβ ) H
.
Gr cosβ
=
Write equations of motion.
Moments about A : + ΣMA= Σ(MA)eff :
W = mg r = 3 in ω = 12 rad/s H
.
G = - 1 mr2ω sinβ cosβ k’2
- mg r sinβ = - (r cos β ) m ω2 (r sinβ ) - mr12 2ω sinβ cosβ β = 53.4o
cos β = =2 g 3 r ω2
2 (32.2) 3 (0.25)(12)2
(a) The constant angle β the ring forms with the vertical.
Problem 18.154 Solution A
β
r sinβ W
Ax Ay
A
G
m a = m ω2 (r sinβ ) H
.
Gr cosβ
=
cos β = 2 g 3 r ω2
(b) The maximum value of β for which the ring remains vertical.
From part (a) : For β = 0o
cos β = 1 = 2 (32.2)
3 (0.25) ω2 ω = 9.27 rad/s