Dynamic Analysis and Forces
For accelerating a mass(m), we need to exert a force on it. Similarly, to cause an
angular acceleration in a rotating body, a torque must be exerted on it (Figure
4.1), as:
To accelerate a robot’s links, it is necessary to have
actuators capable of exerting large enough forces and
torques on the links and joints to move them at a desired
acceleration and velocity.
In general, techniques such as Newtonian mechanics can be used to find the dynamic
equations for robots. However, due to the fact that robots are 3-D and multi-DOF
mechanisms with distributed masses, it is very difficult to use Newtonian mechanics. Instead,
we may use other techniques such as Lagrangian mechanics, which is based on energy
terms only, and therefore, in many cases, easier to use. Although Newtonian mechanics and
other techniques can be used for this derivation, most references are based on Lagrangian
mechanics.
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Lagrangian mechanics is based on the differentiation of the energy terms with respect to
the system’s variables and time. Lagrangian is defined as:
where (L) is the Lagrangian, (K) is the kinetic energy of the system, and (P) is the potential
energy of the system. Then:
where (Fi )is the summation of all external forces for a linear motion,(Ti) is the summation of
all external torques for a rotational motion, and (θi) and (xi) are system variables. As a result,
in order to get the equations of motion, we need to derive energy equations for the
system and then differentiate the Lagrangian according to Equations
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Example
Derive the force-acceleration relationship for the 1-DOF system (schematic of a simple cart-spring
system ) shown below, using both the Lagrangian mechanics as well as the Newtonian mechanics.
Assume the wheels have negligible inertia.
Solution: The x-axis denotes the motion of the cart
and is used as the only variable in this system.
Since this is a 1-DOF system, there will be only one
equation describing the linear motion.
The derivatives of the Lagrangian are:
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Therefore, the equation of motion for the cart will be:
To solve the problem with Newtonian mechanics, we will draw the free-body diagram of the cart (Figure
below) and solve for forces as follows:
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Example 2
Using the Lagrangian method, derive the equations of motion for the 2-DOF robot arm, as
shown in Figure below. The center of mass for each link is at the center of the link. The
moments of inertia are I1 and I2.
Solution: First, we calculate the velocity of the center of mass of link 2 by
differentiating its position:
Therefore, the total velocity of the center of mass of link 2
is
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The total kinetic energy of the system is the sum of the kinetic energies of links 1 and 2.
Substituting Equation The value of (VD ) into the Equation of (K) and regrouping,
we get:
The potential energy of the system is the sum of the potential energies of the two links:
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The Lagrangian for the two-link robot arm will be:
Taking the derivatives of the Lagrangian and substituting the terms into the torque Equation
(Ti)
will yield the following two equations of motion:
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These torque Equations can be written in matrix form as:
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Example: write the dynamic equation for the SCARA robot with the configuration
below
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Lagrangian equation of second order has the following expression
Where:
QK -generalized forces (force F or torque T)
qk -generalized coordinates (displacement d or angle )
L=(Ke –P)- Lagrange function
K = 1, 2, 3, .., n
Ke- kinetic energy
P - Potential energy
For SCARA robot of the figure in the previous slide, kinetic and
potential energies can be written as:
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+ 𝐽2 + 𝑚2 (𝑟22 + 𝐿1 𝑟2 𝑐2 ) 𝜃1 𝜃2
=
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2(𝐽𝑚3 + 𝐽𝑔3
𝑚3 + 𝑚𝑚3
+
𝐷2
2
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𝑑32
11
So, Lagrange function is:
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Then torques exerted on the robot joints can be derived as:
where M denotes the inertia, C, B expresses the Coriolis and
centripetal forces respectively, and G is the gravity vector:
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