Robotics and Control : Theory and Practice - Week 3 Assignment solution

 

1 point

An object of mass m=20 units is dropped from height h=50 units from rest, when it just reaches the ground, its Lagrangian L is (Take g=10):

 5000

 2500

 0

 10000

1 point

If V(x) denotes the Lyapunov function for the system x˙=f(x):f(0)=0$\dot{�}=�(�):�(0)=0$  then:

x=0 is stable if V is positive semi definite and V˙$\dot{�}$ is negative definite.

x=0 is asymptotically stable if V  positive definite and V˙$\dot{�}$ is negative semi-definite.

x=0  is unstable if V is positive definite and V˙$\dot{�}$ is negative definite.

x=0 is stable if V is negative definite and V˙$\dot{�}$ is positive semi-definite.

1 point

Consider following dynamical system:
x1˙=x2$\dot{{�}_1}={�}_2$
x2˙=−sinx1$\dot{{�}_2}=-���{�}_1$
then:

The Lyapunov function (1−cosx1)+x222$(1-���{�}_1)+\frac{{�}_2^2}{2}$ indicates that the system is stable at (0, 0).

The Lyapunov function (1−cosx1)+x222$(1-���{�}_1)+\frac{{�}_2^2}{2}$ indicates that the system is asymptotically stable at (0,0).

The Lyapunov function (1−cosx2)+x212$(1-���{�}_2)+\frac{{�}_1^2}{2}$ indicates that the system is stable at (0,0).

The Lyapunov function (1−sinx1)+x212$(1-���{�}_1)+\frac{{�}_1^2}{2}$ indicates that the system is stable (0,0).

1 point

Torque for a single arm manipulator as shown in figure with length L=1 unit and mass m=2 units at the end when θ=0$�=0$ rad$���$ and θ¨=1$\ddot{�}=1$ rad/s2$���/{�}^2$ is given by (Taking g=10): 

 2

 20

 22

 18

1 point

A point xe∈Rn${�}_{�}\in {�}^{�}$ is said to be an equilibrium point of the system
x˙=f(t,x)$\dot{�}=�(�,�)$

If f(t,xe)=0 for some t.$���(�,{�}_{�})=0��������.$

If f(t,xe)=1 for some t.$���(�,{�}_{�})=1��������.$

If f(t,xe)=0 for all t.$���(�,{�}_{�})=0�������.$

If f(t,xe)=1 for all t.$���(�,{�}_{�})=1�������.$

1 point

Consider following example of a two-arm manipulator with uniformly distributed mass with length l1=3${�}_1=3$ units and l2=4${�}_2=4$ units, moment of inertia I1 and I2${�}_1���{�}_2$ and mass m1=6${�}_1=6$ units and m2=9${�}_2=9$ units for respective links.

then moment of Inertia of link 1 about A and moment of inertia of link 2 about D is given by is:

 36 and 24 respectively.

 24 and 32 respectively.

 18 and 12 respectively.

 24 and 16 respectively.

1 point

Potential energy P2${�}_2$ in question (6) for link 2 is given by when θ1=π/6${�}_1=�/6$  and θ2=0${�}_2=0$ (g=10):

 225

 450

 300

 900

1 point

Kinetic energy K1${�}_1$ in question (6) for link 1 is given by when θ1˙=2$\dot{{�}_1}=2$ is

 36

 16

 48

 128

1 point

For single arm robot manipulator as shown in figure with length m=3 units, and uniformly distributed mass m=5 units, ∂L∂t$\frac{\mathrm{\partial }�}{\mathrm{\partial }�}$ at θ=π/3,θ˙=2$�=�/3,\dot{�}=2$  and θ¨=1$\ddot{�}=1$ unit where L is Lagrangian (Take g=10) is:

 60

 -45

 37.5

 22.5

1 point

Which of the following is true about Lyapunov function V(x) with stable equilibrium point:

 V(x) is not continuous.

 Partial derivatives of V(x) are continuous

V˙(x)$\dot{�}(�)$ must be positive

 V(x) is not a vector valued function