Download Homework 3 Solution

Homework 3 Solution
Problem 1: A speed governor consists of a block of mass
m that slides within a smooth groove in a turntable that
rotates about its center point at angular speed Ω. The
identical opposing springs, whose stiffness is k, are
precompressed. Consequently the springs maintain their
contact with the block regardless of the displacement s.
The system lies in the horizontal plane. Derive an
expression for the normal force exerted by the groove wall
on the block and the differential equation governing s as a
function of time in the case where Ω is an arbitrary
function of time. Then determine the natural frequency of
vibratory motion of the block within the groove for the
case in which Ω is constant, and explain how that result can be used to monitor when Ω exceeds a
critical value.
Solution: Attach a moving coordinate system to the disk, with the moving x axis aligned with the
groove. The origin of the moving coordinate system is at the center of the disk. To solve this
problem we must first find the total acceleration of the block. We will write this acceleration in the
moving coordinate system so that the normal force will be easy to identify.
There is only one angular velocity in this problem
𝑒̂1 = 𝑘̂
𝜔̇ 1 = Ω̇
𝜔1 = Ω
𝛀1 = 𝛚
𝛚 = Ω𝑘̂
𝛂 = 𝜔̇ 1 𝑒̂1 + 𝜔1 (𝛀1 × 𝑒̂1 ) = Ω̇𝑘̂
The vector from the origin to the block is
𝑠
𝐬 = {ℎ }
0
And the velocity and acceleration of the block relative to the moving coordinate system is
𝑠̇
𝐬̇ 𝑟 = {0}
0
𝑠̈
𝐬̈ 𝑟 = {0}
0
Since the origin of the moving coordinate system does not translate, we have
𝐫̈0 = 𝟎
Thus, the total acceleration is
𝑠
𝑠
0
0
0
0
𝑠̈
𝑠̇
𝐚 = {0} + { 0 } × {ℎ} + 2 { 0 } × {0} + { 0 } × ({ 0 } × {ℎ})
0
0
Ω
Ω
Ω
Ω̇
0
0
𝑠
0
−ℎ
𝑠̈
2 ℎ
̇
𝐚 = {0} + Ω { 𝑠 } + {2𝑠̇ Ω} − Ω { }
0
0
0
0
𝑠̈ − Ω̇ℎ − Ω2 𝑠
𝐚 = {Ω̇𝑠 + 2𝑠̇ Ω − Ω2 ℎ}
0
Use the 𝑗̂ component to find the normal force on the sidewall of the groove:
𝐹𝑛 = 𝑚 (Ω̇ 𝑠 + 2𝑠̇ Ω − Ω2 ℎ)
The equation of motion along the groove is
∑ 𝐹𝑡 = −2𝑘𝑠 = 𝑚 (𝑠̈ − Ω̇ ℎ − Ω2 𝑠)
Rearrange this into standard form
2
𝑚𝑠̈ + (2𝑘 − 𝑚Ω ) 𝑠 = 𝑚Ω̇ ℎ
If the angular velocity of the disk is constant, this has the form of a mass/spring system
𝑚𝑥̈ + 𝑘𝑥 = 0
And the natural frequency is
2𝑘 − 𝑚Ω2
2𝑘
√
𝜔𝑛 =
= √ − Ω2
𝑚
𝑚
If
Ω2 >
2𝑘
𝑚
then the natural frequency will be imaginary, which implies an unstable system. In reality, the mass
will cease to vibrate back and forth, and will remain pinned against one end of the groove. If a limit
switch were placed at the end of the groove, it could be used to indicate when the disk had reached a
critical speed.
Problem 2: The disk spins about its axis CD at 𝜑̇ as
the system rotates about the vertical axis at 𝜃̇. The
angle of elevation of the arm supporting the disk is β.
The rates 𝜑̇ , 𝜃̇ , 𝛽̇ are constant. The length from A to B
and from B to the axis of rotation of the disk is L, and
the radius of the disk is R. Determine the velocity and
acceleration of point E, which is the lowest point on
the perimeter of the disk.
Solution: There are several ways we could go about
solving this problem. In class, we have typically
attached the moving coordinate system to the object
with the most “motion” – in this case, the disk. We
will try a slightly different approach this time by attaching the moving coordinate system to the outer
arm, with its origin at B. The moving x axis is aligned with the arm and the moving z axis is parallel
to the axis of rotation of the disk. To get from the fixed coordinate system to the moving
coordinate system requires one rotation about the Y axis.
cos 𝛽
𝐑𝑦 = [ 0
sin 𝛽
cos 𝛽
𝐼̂𝑥𝑦𝑧 = [ 0
sin 𝛽
0
1
0
− sin 𝛽 1
cos 𝛽
0 ] {0} = { 0 }
cos 𝛽 0
sin 𝛽
0 − sin 𝛽
1
0 ]
0 cos 𝛽
cos 𝛽
̂𝑥𝑦𝑧 = [ 0
𝐾
sin 𝛽
0
1
0
− sin 𝛽 0
− sin 𝛽
0 ] {0} = { 0 }
cos 𝛽 1
cos 𝛽
There are two angular velocities associated with this system
𝜔1 = 𝜃̇
𝜔2 = 𝛽̇
̂
𝑒̂1 = 𝐾
𝑒̂2 = 𝑗̂
𝜔̇ 1 = 0
𝜔̇ 2 = 0
The total angular velocity of the outer arm is
−𝜃̇ sin 𝛽
𝛚 = { 𝛽̇
}
𝜃̇ cos 𝛽
The angular acceleration of the outer arm is
𝛂 = 𝜔̇ 1 𝑒̂1 + 𝜔1 (𝛀1 × 𝑒̂1 ) + 𝜔̇ 2 𝑒̂2 + 𝜔2 (𝛀2 × 𝑒̂2 )
−𝛽̇ 𝜃̇ cos 𝛽
𝛂={
}
0
−𝛽̇ 𝜃̇ sin 𝛽
The vector from the origin of the moving coordinate system to the point E is
𝛀1 = 0
𝛀2 = 𝛚
𝐿+𝑅
𝐬={ 0 }
0
From the viewpoint of an observer in the moving coordinate system, the velocity of point E is
purely tangential arising from the rotation of the disk.
0
𝐬̇ 𝑟 = {𝑅𝜑̇ }
0
and the relative acceleration of point E is entirely centripetal
−𝑅𝜑̇ 2
𝐬̈ 𝑟 = { 0 }
0
The acceleration at point B is also purely centripetal
−𝐿𝜃̇ 2 cos 𝛽
𝐫̈0 = −𝐿𝜃 𝐼 = {
}
0
−𝐿𝜃̇ 2 sin 𝛽
̇2 ̂
Some of the necessary cross-products are
−(𝐿 + 𝑅)𝛽̇ 2 − (𝐿 + 𝑅)𝜃̇ 2 cos2 𝛽
𝛚 × (𝛚 × 𝐬) = {
}
−(𝐿 + 𝑅)𝛽̇𝜃̇ sin 𝛽
2
̇
−(𝐿 + 𝑅)𝜃 cos 𝛽 sin 𝛽
0
𝛂 × 𝐬 = {−(𝐿 + 𝑅)𝛽̇ 𝜃̇ sin 𝛽}
0
−2𝑅𝜑̇ 𝜃̇ cos 𝛽
2𝛚 × 𝐬̇ 𝑟 = {
}
0
−2𝑅𝜑̇ 𝜃̇ sin 𝛽
Thus, the total acceleration at point E is
−𝑅𝜑̇ 2 − 𝐿𝜃̇ 2 cos 𝛽 − (𝐿 + 𝑅)𝛽̇ 2 − (𝐿 + 𝑅)𝜃̇ 2 cos2 𝛽 − 2𝑅𝜑̇ 𝜃̇ cos 𝛽
𝐚𝐸 = {
}
−2(𝐿 + 𝑅)𝛽̇ 𝜃̇ sin 𝛽
2
2
̇
̇
̇
−𝐿𝜃 sin 𝛽 − (𝐿 + 𝑅)𝜃 cos 𝛽 sin 𝛽 − 2𝑅𝜑̇ 𝜃 sin 𝛽
Problem 3: Collar C slides relative to the curved rod at a constant speed u. The rotation rate about
bearing axis AB is constant at Ω. Determine the acceleration of the collar in terms of the angle θ.
Also derive expressions for the dynamic forces exerted on the collar by the rod and the tangential
force required to hold u constant. Gravity may be neglected.
Solution: Attach a moving coordinate system to the center of the ring with the –y axis aligned with
the collar, as shown in the figure above. There is one rotation needed to switch from the fixed XYZ
system to the moving xyz system
cos 𝜃
𝐑 𝑧 = [− sin 𝜃
0
sin 𝜃
cos 𝜃
0
0
0]
1
cos 𝜃
𝐼̂𝑥𝑦𝑧 = {− sin 𝜃 }
0
sin 𝜃
𝐽̂𝑥𝑦𝑧 = {cos 𝜃 }
0
There are two angular velocities associated with this system
𝜔1 = Ω
𝜔2 = 𝜃̇
𝑒̂1 = 𝐼̂
𝑒̂2 = 𝑘̂
𝜔̇ 1 = 0
𝜔̇ 2 = 0
But the rate of change of θ can be written in terms of the velocity of the collar
𝜃̇ =
𝑢
𝑅
Thus, the total angular velocity is
Ω cos 𝜃
𝛚 = {−Ω sin 𝜃}
𝑢/𝑅
𝛀1 = 0
𝛀2 = 𝛚
And the angular acceleration is
𝛂=−
𝑢Ω sin 𝜃
{cos 𝜃}
𝑅
0
The vector from the origin of the moving coordinate system to the collar is
0
𝐬 = {−𝑅}
0
𝐬̇ 𝑟 = 𝟎
𝐬̈ 𝑟 = 𝟎
The acceleration of the origin of the moving coordinate system is entirely centripetal
sin 𝜃
𝐫̈0 = −𝑅Ω 𝐽 = −𝑅Ω {cos 𝜃}
0
2̂
2
A few of the necessary cross products are
𝑅Ω2 sin 𝜃 cos 𝜃
𝛚 × (𝛚 × 𝐬) = {𝑅Ω2 cos 2 𝜃 + 𝑢2 /𝑅 }
𝑢Ω sin 𝜃
0
𝛂×𝐬= { 0 }
𝑢Ω sin 𝜃
And the total acceleration of the collar is
−𝑅Ω2 sin 𝜃 (1 − cos 𝜃)
𝐚 = {−𝑅Ω2 cos 𝜃 (1 − cos 𝜃) + 𝑢2 /𝑅}
2𝑢Ω sin 𝜃
The tangential force is the mass multiplied by the x component of the acceleration. The other two
components (multiplied by mass) give the force that is exerted by the ring on the collar.