Download Analytical Dynamics Homework 7

Analytical Dynamics Homework 7 - Solution
Use Lagrange’s Equations to determine the equations of motion for the following systems.
Problem 1: The cart with mass 4m rolls without friction on the ground and is attached to the wall
with a spring of stiffness 2k. The cart forms a wedge with interior angle θ. On top of the cart is a
smaller mass which is attached with a spring of stiffness k. Determine the equations of motion for
the system.
Solution: This is a 2DOF problem, and we choose as generalized coordinates the horizontal
displacement of the cart, x, and the movement of the block along the top of the cart, s. The
movements of the cart and block are given by
𝑥
𝐫𝑐 = { }
0
𝑥 + 𝑠 cos 𝜃
𝐫𝑏 = {
}
−𝑠 sin 𝜃
The velocities of each mass are
𝑥̇
𝐫̇𝑐 = { }
0
𝑥̇ + 𝑠̇ cos 𝜃
𝐫̇𝑏 = {
}
−𝑠 sin 𝜃
The kinetic energy of the cart is
𝑇𝑐 =
1
(4𝑚)𝐫̇𝑐 ⋅ 𝐫̇𝑐 = 2𝑚𝑥̇ 2
2
And the kinetic energy of the block is
𝑇𝑏 =
1
1
𝑚𝐫̇𝑏 ⋅ 𝐫̇𝑏 = 𝑚(𝑥̇ 2 + 2𝑥̇ 𝑠̇ cos 𝜃 + 𝑠̇ 2 )
2
2
Thus, the total kinetic energy is
1
𝑇=
1
𝑚(5𝑥̇ 2 + 2𝑥̇ 𝑠̇ cos 𝜃 + 𝑠̇ 2 )
2
The potential energy is provided by the springs and gravity:
1
1
𝑉 = (2𝑘)𝑥 2 + 𝑘𝑠 2 − 𝑚𝑔𝑠 sin 𝜃
2
2
Now take the appropriate derivatives
𝜕𝑇
= 𝑚(5𝑥̇ + 𝑠̇ cos 𝜃)
𝜕𝑥̇
𝜕𝑇
= 𝑚(𝑠̇ + 𝑥̇ cos 𝜃)
𝜕𝑠̇
𝑑 𝜕𝑇
( ) = 𝑚(5𝑥̈ + 𝑠̈ cos 𝜃)
𝑑𝑡 𝜕𝑥̇
𝑑 𝜕𝑇
( ) = 𝑚(𝑠̈ + 𝑥̈ cos 𝜃)
𝑑𝑡 𝜕𝑠̇
𝜕𝑇
=0
𝜕𝑥
𝜕𝑇
=0
𝜕𝑠
𝜕𝑉
= 2𝑘𝑥
𝜕𝑥
𝜕𝑉
= 𝑘𝑠 − 𝑚𝑔 sin 𝜃
𝜕𝑠
Thus, the equations of motion for the system are
𝑚(5𝑥̈ + 𝑠̈ cos 𝜃) + 2𝑘𝑥 = 0
𝑚(𝑠̈ + 𝑥̈ cos 𝜃) + 𝑘𝑠 = 𝑚𝑔 sin 𝜃
Or, in matrix form
[
5
cos 𝜃
2𝑘𝑥⁄
cos 𝜃 𝑥̈
𝑚
]{ } = {
}
𝑘𝑠⁄ + 𝑔 sin 𝜃
𝑠̈
1
𝑚
2
Problem 2: Each link in the double pendulum has mass m and length l. Determine the equations of
motion for the pendulum if an applied horizontal force F(t) acts at point C.
Solution: The moment of inertia of each thin rod is
𝐼=
𝑚𝑙 2
12
This is also a 2DOF problem, and we choose θ1 and θ2 as generalized coordinates. First, find a
vector to the point of application of the force
𝐫𝐶 = 𝑙 {
cos 𝜃1 + cos 𝜃2
}
sin 𝜃1 + sin 𝜃2
The virtual displacement at point C in terms of the generalized coordinates is
𝛿𝑟𝐶 = 𝑙 {
− sin 𝜃1
− sin 𝜃2
} 𝛿𝜃1 + 𝑙 {
} 𝛿𝜃2
cos 𝜃1
cos 𝜃2
The force at point C is entirely horizontal, so that
𝐹
𝐅={ }
0
The virtual work performed by this force is then
𝛿𝑊 = −𝐹𝑙 sin 𝜃1 𝛿𝜃1 − 𝐹𝑙 sin 𝜃2 𝛿𝜃2
Thus, the generalized forces are
𝑄1 = −𝐹𝑙 sin 𝜃1
𝑄2 = −𝐹𝑙 sin 𝜃2
3
Remember that the angles shown in the figure are negative, so the generalized forces will create positive
moments for each generalized coordinate. The locations of the centers of mass are
𝑙 cos 𝜃1
𝐫1 = {
}
2 sin 𝜃1
𝐫2 = 𝑙 {
𝑙 cos 𝜃2
cos 𝜃1
}+ {
}
sin 𝜃1
2 sin 𝜃2
The velocities of the centers of mass are
𝐫̇1 =
𝑙𝜃̇1 − sin 𝜃1
{
}
2 cos 𝜃1
𝐫̇2 = 𝑙𝜃̇1 {
𝑙𝜃̇2 − sin 𝜃2
− sin 𝜃1
}+
{
}
cos 𝜃1
2 cos 𝜃2
The kinetic energy of the top pendulum is
1
1 2 𝑚𝑙 2 2 𝑚𝑙 2 2 𝑚𝑙 2 2
𝑇1 = 𝑚𝐫̇1 ⋅ 𝐫̇1 + 𝐼𝜃̇1 =
𝜃̇ +
𝜃̇ =
𝜃̇
2
2
8 1
24 1
6 1
The kinetic energy of the bottom pendulum is
𝑇2 =
1
1
𝑚𝐫̇2 ⋅ 𝐫̇2 + 𝐼𝜃̇22
2
2
1
𝑙2
𝑚𝑙 2 2
𝑇2 = 𝑚 (𝑙 2 𝜃̇12 + 𝑙 2 𝜃̇1 𝜃̇2 (sin 𝜃1 sin 𝜃2 + cos 𝜃1 cos 𝜃2 ) + 𝜃̇22 ) +
𝜃̇
2
4
24 2
𝑇2 =
𝑚𝑙 2 2 𝑚𝑙 2
𝑚𝑙 2 2
𝜃̇1 +
𝜃̇1 𝜃̇2 cos(𝜃2 − 𝜃1 ) +
𝜃̇
2
2
6 2
Thus, the total kinetic energy is
𝑚𝑙 2 4 2 1 2
𝑇=
( 𝜃̇ + 𝜃̇ + 𝜃̇1 𝜃̇2 cos(𝜃2 − 𝜃1 ))
2 3 1 3 2
The potential energy is
𝑉=
𝑚𝑔𝑙
𝑚𝑔𝑙
sin 𝜃1 + 𝑚𝑔𝑙 sin 𝜃1 +
sin 𝜃2
2
2
𝑉=
3𝑚𝑔𝑙
𝑚𝑔𝑙
sin 𝜃1 +
sin 𝜃2
2
2
Now take the appropriate derivatives
𝜕𝑇
𝑚𝑙 2 8
=
( 𝜃̇ + 𝜃̇2 cos(𝜃2 − 𝜃1 ))
2 3 1
𝜕𝜃̇1
𝜕𝑇
𝑚𝑙 2 2
=
( 𝜃̇ + 𝜃̇1 cos(𝜃2 − 𝜃1 ))
2 3 2
𝜕𝜃̇2
4
𝑑 𝜕𝑇
𝑚𝑙 2 8
( 𝜃̈ + 𝜃̈2 cos(𝜃2 − 𝜃1 ) − 𝜃̇2 (𝜃̇2 − 𝜃̇1 ) sin(𝜃2 − 𝜃1 ))
(
)=
𝑑𝑡 𝜕𝜃̇1
2 3 1
𝑑 𝜕𝑇
𝑚𝑙 2 2
( 𝜃̈ + 𝜃̈1 cos(𝜃2 − 𝜃1 ) − 𝜃̇1 (𝜃̇2 − 𝜃̇1 ) sin(𝜃2 − 𝜃1 ))
(
)=
𝑑𝑡 𝜕𝜃̇2
2 3 2
𝜕𝑇
𝑚𝑙 2
=
𝜃̇ 𝜃̇ sin(𝜃2 − 𝜃1 )
𝜕𝜃1
2 1 2
𝜕𝑇
𝑚𝑙 2
=−
𝜃̇ 𝜃̇ sin(𝜃2 − 𝜃1 )
𝜕𝜃2
2 1 2
𝜕𝑉
3𝑚𝑔𝑙
=
cos 𝜃1
𝜕𝜃1
2
𝜕𝑉
𝑚𝑔𝑙
=
cos 𝜃2
𝜕𝜃2
2
Finally, the equations of motion are:
𝑚𝑙 2 8
3
[ 𝜃̈ 1 + 𝜃̈ 2 cos(𝜃2 − 𝜃1 ) − 𝜃̇22 sin(𝜃2 − 𝜃1 )] + 𝑚𝑔𝑙 cos 𝜃1 = −𝐹𝑙 sin 𝜃1
2 3
2
𝑚𝑙 2 2 ̈
1
2
[ 𝜃2 + 𝜃̈ 1 cos(𝜃2 − 𝜃1 ) + 𝜃̇ 1 sin(𝜃2 − 𝜃1 )] + 𝑚𝑔𝑙 cos 𝜃2 = −𝐹𝑙 sin 𝜃2
2 3
2
Or, in slightly simplified form
8 ̈
3𝑔
2𝐹
𝜃 + 𝜃̈ 2 cos(𝜃2 − 𝜃1 ) − 𝜃̇22 sin(𝜃2 − 𝜃1 ) +
cos 𝜃1 = − sin 𝜃1
3 1
𝑙
𝑚𝑙
2 ̈
𝑔
2𝐹
2
𝜃2 + 𝜃̈ 1 cos(𝜃2 − 𝜃1 ) + 𝜃̇ 1 sin(𝜃2 − 𝜃1 ) + cos 𝜃2 = − sin 𝜃2
3
𝑙
𝑚𝑙
5
Problem 3: The cylinder of radius r and mass m rolls without slipping in the cylindrical cavity in the
cart, which has radius R. The cart, which has mass M, is attached to the wall with a spring of
stiffness k. An applied force, F(t) acts on the cart in the horizontal direction. Neglecting friction,
what are the equations of motion for the system?
Solution: This is also a 2DOF problem, and we choose x and θ as generalized coordinates. The
cylinder has moment of inertia
𝑚𝑟 2
𝐼=
2
Since the cylinder rolls without slipping, the velocity of its center point is the same as the tangential
velocity of the rolling cylinder. Let us define
𝜌 =𝑅−𝑟
Then the velocity of the center of the cylinder is
𝑣𝐶 = 𝜌𝜃̇
If we define the angular velocity of the cylinder to be 𝜑̇ , then the tangential velocity of the cylinder
is
𝑣𝐶 = 𝑟𝜑̇
Thus,
𝜑̇ =
𝜌
𝜃̇
𝑟
The displacement and velocity of the cart is
𝑥
𝐫𝑊 = { }
0
𝑥̇
𝐫̇𝑊 = { }
0
6
Assuming that the “neutral” position of the cylinder is at the bottom of the cart, when θ = 0, the
displacement and velocity of the cylinder is
𝑥 − 𝜌 sin 𝜃
𝐫𝐶 = {
}
𝜌(1 − cos 𝜃)
𝑥̇ − 𝜌𝜃̇ cos 𝜃
𝐫̇𝐶 = {
}
𝜌𝜃̇ sin 𝜃
The kinetic energy of the cart is
1
𝑇𝑊 = 𝑀𝑥̇ 2
2
and the kinetic energy of the cylinder is
𝑇𝐶 =
1
1
𝑚𝐫̇𝐶 ⋅ 𝐫̇𝐶 + 𝐼𝜑̇ 2
2
2
1
1 𝑚𝑟 2 𝜌2 2
𝑇𝐶 = 𝑚(𝑥̇ 2 − 2𝜌𝑥̇ 𝜃̇ cos 𝜃 + 𝜌2 𝜃̇ 2 ) + (
) 𝜃̇
2
2 2 𝑟2
𝑇𝐶 =
1
3
𝑚(𝑥̇ 2 − 2𝜌𝑥̇ 𝜃̇ cos 𝜃 + 𝜌2 𝜃̇ 2 )
2
2
The total kinetic energy is
𝑇=
1
3
(𝑀 + 𝑚)𝑥̇ 2 − 𝑚𝜌𝑥̇ 𝜃̇ cos 𝜃 + 𝑚𝜌2 𝜃̇ 2
2
4
The potential energy is
1
𝑉 = 𝑘𝑥 2 + 𝑚𝑔𝜌(1 − cos 𝜃)
2
Now take the appropriate derivatives
𝜕𝑇
= (𝑀 + 𝑚)𝑥̇ − 𝑚𝜌𝜃̇ cos 𝜃
𝜕𝑥̇
𝜕𝑇
3
= −𝑚𝜌𝑥̇ cos 𝜃 + 𝑚𝜌2 𝜃̇
2
𝜕𝜃̇
𝑑 𝜕𝑇
( ) = (𝑀 + 𝑚)𝑥̈ − 𝑚𝜌𝜃̈ cos 𝜃 + 𝑚𝜌𝜃̇ 2 sin 𝜃
𝑑𝑡 𝜕𝑥̇
𝑑 𝜕𝑇
3
( ) = −𝑚𝜌𝑥̈ cos 𝜃 + 𝑚𝜌𝜃̇ 𝑥̇ sin 𝜃 + 𝑚𝜌2 𝜃̈
̇
𝑑𝑡 𝜕𝜃
2
7
𝜕𝑇
=0
𝜕𝑥
𝜕𝑇
= 𝑚𝜌𝑥̇ 𝜃̇ sin 𝜃
𝜕𝜃
𝜕𝑉
= 𝑘𝑥
𝜕𝑥
𝜕𝑉
= 𝑚𝑔𝜌 sin 𝜃
𝜕𝜃
Collect all terms to form the equations of motion
2
(𝑀 + 𝑚)𝑥̈ − 𝑚𝜌𝜃̈ cos 𝜃 + 𝑚𝜌𝜃̇ sin 𝜃 + 𝑘𝑥 = 𝐹
3
𝑚𝜌2 𝜃̈ − 𝑚𝜌𝑥̈ cos 𝜃 + 𝑚𝑔𝜌 sin 𝜃 = 0
2
8