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### Week 01: Modern Robotics, Course 3: Robot Dynamics Coursera Quiz Answers

#### Quiz 01: Lecture Comprehension, Lagrangian Formulation of Dynamics (Chapter 8 through 8.1.2, Part 1 of 2)

Q1. The Lagrangian for a mechanical system is

**the kinetic energy plus the potential energy.**- the kinetic energy minus the potential energy.

Q2. To evaluate the Lagrangian equations of motion,

\tau_i = \frac{d}{dt} \frac{\partial \mathcal{L}}{\partial \dot{\theta}_i} – \frac{\partial \mathcal{L}}{\partial \theta_i}*τi*=*dtd*∂*θ*˙*i*∂L−∂*θi*∂L,

you must be able to take derivatives, such as the partial derivative with respect to a joint variable or velocity or a total derivative with respect to time. Therefore, the product rule and chain rules for derivatives will be useful. (If you have forgotten them, you can refresh your memory with any standard reference, including Wikipedia.) Which of the answers below represents the time derivative of 2 \theta_1 \cos(4 \theta_2)2*θ*1cos(4*θ*2), where \theta_1*θ*1 and \theta_2*θ*2 are functions of time?

- -8 \dot{\theta}_1 \sin(4 \theta_2)−8
*θ*˙1sin(4*θ*2) **2 \dot{\theta}_1 \cos (4 \theta_2) – 8 \theta_1 \sin(4 \theta_2) \dot{\theta}_22***θ*˙1cos(4*θ*2)−8*θ*1sin(4*θ*2)*θ*˙2- 2 \dot{\theta}_1 \cos (4 \theta_2) – 2 \theta_1 \sin(4 \theta_2) \dot{\theta}_22
*θ*˙1cos(4*θ*2)−2*θ*1sin(4*θ*2)*θ*˙2 - 2 \dot{\theta}_1 \cos (4 \theta_2) + 2 \theta_1 \cos(4 \theta_2) \dot{\theta}_22
*θ*˙1cos(4*θ*2)+2*θ*1cos(4*θ*2)*θ*˙2

Q3. The equations of motion for a robot can be summarized as

\tau = M(\theta) \ddot{\theta} + c(\theta,\dot{\theta}) + g(\theta)*τ*=*M*(*θ*)*θ*¨+*c*(*θ*,*θ*˙)+*g*(*θ*).

If the equation for the first joint is

\tau_1 = term1 + term2 + term3 + …*τ*1=*term*1+*term*2+*term*3+…

Q4. which of the following terms, written in terms of (\theta,\dot{\theta},\ddot{\theta})(*θ*,*θ*˙,*θ*¨), could **not** be one of the terms on the right-hand side of the equation? (The value k*k* is not a function of (\theta,\dot{\theta},\ddot{\theta})(*θ*,*θ*˙,*θ*¨) and could represent constants like link lengths, masses, or inertias, as needed to get correct units.) Select all that apply.

- k\ddot{\theta}_2 \cos(\theta_1)
*kθ*¨2cos(*θ*1) **k\ddot{\theta}_1 \dot{\theta}_1***kθ*¨1*θ*˙1- k\sin \theta_3
*k*sin*θ*3 - k\dot{\theta}_1 \dot{\theta}_2 \sin \theta_2
*kθ*˙1*θ*˙2sin*θ*2 **k \dot{\theta}_1 \sin \theta_2***kθ*˙1sin*θ*2

### Quiz 02: Lecture Comprehension, Lagrangian Formulation of Dynamics (Chapter 8 through 8.1.2, Part 2 of 2)

Q1. Which of the following could be a centripetal term in the dynamics?

**k \dot{\theta}_1^2***kθ*˙12- k \dot{\theta}_1 \dot{\theta}_2
*kθ*˙1*θ*˙2

Q2. Which of the following could be a Coriolis term in the dynamics?

- k \dot{\theta}_1^2
*kθ*˙12 **k \dot{\theta}_1 \dot{\theta}_2***kθ*˙1*θ*˙2

Q3. One form of the equations of motion is

\tau = M(\theta)\ddot{\theta} + \dot{\theta}^{\rm T} \Gamma(\theta) \dot{\theta} + g(\theta).*τ*=*M*(*θ*)*θ*¨+*θ*˙TΓ(*θ*)*θ*˙+*g*(*θ*).

Which of the following is true about \Gamma(\theta)Γ(*θ*)? Select all that apply.

**\Gamma(\theta)Γ(***θ*) is zero if the mass matrix M*M*has no dependence on \theta*θ*.- \Gamma(\theta)Γ(
*θ*) is an n \times n*n*×*n*matrix, where n*n*is the number of joints. - \Gamma(\theta)Γ(
*θ*) depends on M(\theta)*M*(*θ*) and \dot{\theta}*θ*˙.

#### Lecture Comprehension, Understanding the Mass Matrix (Chapter 8.1.3)Quiz 03:

Q1. Which of these is a possible mass matrix M(\theta)*M*(*θ*) for a two-joint robot at a particular configuration \theta*θ*? Select all that apply.

- \left[
- 200−1
- \right][200−1]
- \left[
- 4123
- \right][4123]
- \left[
- 3112
**\right][3112]**- \left[
- 2221
- \right][2221]

Q2. True or false? If you grab the end-effector of a robot and try to move it around by hand, the apparent mass (to you) depends on the configuration of the robot.

**True.**- False.

Q3. True or false? If you apply (by hand) a linear force to the end-effector of a robot, the end-effector will accelerate in the same direction as the applied force.

- Always true.
- Always false.
**Sometimes true, sometimes false.**

#### Quiz 04: Lecture Comprehension, Dynamics of a Single Rigid Body (Chapter 8.2, Part 1 of 2)

Q1. How is the center of mass of a rigid body defined?

- The point at the geometric centroid of the body.
**The point at the mass-weighted (or density-weighted) centroid of the body.**

Q2. If the body consists of a set of rigidly connected point masses, with a frame {b} at the center of mass, what is the wrench \mathcal{F}_bF*b* needed to generate the acceleration \dot{{\mathcal V}}_bV˙*b* when the body’s current twist is \mathcal{V}_bV*b*?

- The acceleration \dot{{\mathcal V}}_bV˙
*b* defines the linear acceleration of each point mass in the inertial frame {b}. The linear component f_b*fb* of \mathcal{F}_bF*b* is the sum of the individual vector forces needed to cause those point-mass accelerations (using f=ma*f*=*ma*), and the moment m_b*mb* is the sum of the moments the individual linear forces create in {b}. **Together, \mathcal{V}_bV***b* and \dot{{\mathcal V}}_bV˙*b* define the linear acceleration of each point mass in the inertial frame {b}. The linear component f_b*fb* of \mathcal{F}_bF*b* is the sum of the individual vector forces needed to cause those point-mass accelerations (using f=ma*f*=*ma*), and the moment m_b*mb* is the sum of the moments the individual linear forces create in {b}.

Q3. What is the kinetic energy of a rotating rigid body?

**\frac{1}{2} \omega_b^{\rm T} \mathcal{I}_b \omega_b21***ωb*TI*b**ωb*- \frac{1}{2} \mathcal{I}_b \omega_b^221I
*b**ωb*2

Q4. True or false? For a given body, there is exactly one orientation of a frame at the center of mass that yields a diagonal rotational inertia matrix.

- True.
**False.**

#### Quiz 05: Lecture Comprehension, Dynamics of a Single Rigid Body (Chapter 8.2, Part 2 of 2)

Q1. The spatial inertia matrix is the 6×6 matrix

\mathcal{G}_B = \left[

I*b*00m*I*

\right]G*B*=[I*b*00m*I*].

What is the maximum number n*n* of **unique** nonzero values the spatial inertia could have? In other words, even though the 6×6 matrix has 36 entries, we only need to store n*n* numbers to represent the spatial inertia matrix.

- 6
**7**- 10
- 12

Q2. The expression [\omega_1][\omega_2]-[\omega_2][\omega_1[*ω*1][*ω*2]−[*ω*2][*ω*1 is the so(3)*so*(3) 3×3 skew-symmetric matrix representation of the cross product of two angular velocities, \omega_1 \times \omega_2 = [\omega_1]\omega_2 \in \mathbb{R}^3*ω*1×*ω*2=[*ω*1]*ω*2∈R3. An analogous expression for twists is [\mathcal{V}_1][\mathcal{V}_2]-[\mathcal{V}_2][\mathcal{V}_1][V1][V2]−[V2][V1], the 4×4 se(3)*se*(3) representation of the **Lie bracket** of \mathcal{V}_1V1 and \mathcal{V}_2V2, sometimes written [{\rm ad}_{\mathcal{V}_1}] \mathcal{V}_2 \in \mathbb{R}^6[adV1]V2∈R6. Which of the following statements is true? Select all that apply.if(typeof ez_ad_units!=’undefined’){ez_ad_units.push([[580,400],’networkingfunda_com-medrectangle-4′,’ezslot_4′,340,’0′,’0′])};__ez_fad_position(‘div-gpt-ad-networkingfunda_com-medrectangle-4-0’);

- The matrix [{\rm ad}_{\mathcal{V}_1}][adV1] is an element of se(3)
*se*(3). **[{\rm ad}_{\mathcal{V}_1}] \mathcal{V}_2 = -[{\rm ad}_{\mathcal{V}_2}] \mathcal{V}_1[adV1]V2=−[adV2]V1**

Q3. The dynamics of a rigid body, in a frame at the center of mass {b}, can be written

\mathcal{F}_b = \mathcal{G}_b \dot{\mathcal{V}}_b – [{\rm ad}_{\mathcal{V}_b}]^{\rm T} \mathcal{G}_b \mathcal{V}_bF*b*=G*b*V˙*b*−[adV*b*]TG*b*V*b*.

If \mathcal{V}_b = (\omega_b,v_b) = (0,v_b)V*b*=(*ωb*,*vb*)=(0,*vb*) and \dot{\mathcal{V}}_b = (\dot{\omega}_b, \dot{v}_b) = (0,0)V˙*b*=(*ω*˙*b*,*v*˙*b*)=(0,0), which of the following is true?

**\mathcal{F}_bF***b* is zero.- \mathcal{F}_bF
*b* is nonzero. - Either of the above could be true.

Q4. The dynamics of a rigid body, in a frame at the center of mass {b}, can be written

\mathcal{F}_b = \mathcal{G}_b \dot{\mathcal{V}}_b – [{\rm ad}_{\mathcal{V}_b}]^{\rm T} \mathcal{G}_b \mathcal{V}_bF*b*=G*b*V˙*b*−[adV*b*]TG*b*V*b*.

If \mathcal{V}_b = (\omega_b,v_b) = (\omega_b,0)V*b*=(*ωb*,*vb*)=(*ωb*,0) and \dot{\mathcal{V}}_b = (\dot{\omega}_b, \dot{v}_b) = (0,0)V˙*b*=(*ω*˙*b*,*v*˙*b*)=(0,0), which of the following is true?

- \mathcal{F}_bF
*b* is zero. - \mathcal{F}_bF
*b* is nonzero. **Either of the above could be true.**

#### Quiz 06: Chapter 8 through 8.3, Dynamics of Open Chains

Q1. Consider an iron dumbbell consisting of a cylinder connecting two solid spheres at either end of the cylinder. The density of the dumbbell is 5600 kg/m^33. The cylinder has a diameter of 4 cm and a length of 20 cm. Each sphere has a diameter of 20 cm. Find the approximate rotational inertia matrix \mathcal{I}_bI*b* in a frame {b} at the center of mass with z-axis aligned with the length of the dumbbell. Your entries should be written in units of kg-m^2, and the maximum allowable error for any matrix entry is 0.01, so give enough decimal places where necessary.

**Write the matrix in the answer box and click “Run”:**

[[1.11,2.22,3.33],[4.44,5.55,6.66],[7.77,8.88,9.99]] for \left[

1.114.447.772.225.558.883.336.669.99

\right]⎣⎢⎡1.114.447.772.225.558.883.336.669.99⎦⎥⎤.

- 1
- [[0,0,0],[0,0,0],[0,0,0]]

Q2. The equations of motion for a particular 2R robot arm can be written M(\theta)\ddot{\theta} + c(\theta,\dot{\theta}) + g(\theta) = \tau*M*(*θ*)*θ*¨+*c*(*θ*,*θ*˙)+*g*(*θ*)=*τ*. The Lagrangian \mathcal{L}(\theta,\dot{\theta})L(*θ*,*θ*˙) for the robot can be written in components as

\mathcal{L}(\theta,\dot{\theta}) = \mathcal{L}^1(\theta,\dot{\theta}) + \mathcal{L}^2 (\theta,\dot{\theta}) + \mathcal{L}^3 (\theta,\dot{\theta}) +\ldotsL(*θ*,*θ*˙)=L1(*θ*,*θ*˙)+L2(*θ*,*θ*˙)+L3(*θ*,*θ*˙)+…

One of these components is \mathcal{L}^1 = \mathfrak{m} \dot{\theta}_1 \dot{\theta}_2 \sin\theta_2 L1=m*θ*˙1*θ*˙2sin*θ*2.

Find the right expression for the component of the joint torque \tau^1_1*τ*11 at joint 1 corresponding to the component \mathcal{L}^1L1.

- \tau^1_1 = \mathfrak{m} \ddot{\theta_2} \sin \theta_2 – \mathfrak{m} \dot \theta_2^2 \cos \theta_2
*τ*11=m*θ*2¨sin*θ*2−m*θ*˙22cos*θ*2 - \tau^1_1 = \mathfrak{m} \ddot{\theta_2} \sin \theta_2 + \mathfrak{m} \dot \theta_2^2 \cos \theta_2
*τ*11=m*θ*2¨sin*θ*2+m*θ*˙22cos*θ*2 - \tau^1_1 = \mathfrak{m} \ddot{\theta_2} \cos \theta_2 + \mathfrak{m} \dot \theta_2^2 \sin \theta_2
*τ*11=m*θ*2¨cos*θ*2+m*θ*˙22sin*θ*2

Q3. Referring back to Question 2, find the right expression for the component of joint torque \tau^1_2*τ*21 at joint 2 corresponding to the component \mathcal{L}^1L1.

- \tau^1_2 = \mathfrak{m} \ddot{\theta_2} \sin \theta_2 + \mathfrak{m} \dot \theta_2^2 \cos \theta_2
*τ*21=m*θ*2¨sin*θ*2+m*θ*˙22cos*θ*2 - \tau^1_2 = \mathfrak{m} \ddot{\theta_1} \sin \theta_2
*τ*21=m*θ*1¨sin*θ*2 - \tau^1_2 = \mathfrak{m} \ddot{\theta_1} \sin \theta_2 + \mathfrak{m} \dot \theta_1 \dot \theta_2 \cos \theta_2
*τ*21=m*θ*1¨sin*θ*2+m*θ*˙1*θ*˙2cos*θ*2

Q4. For a given configuration \theta*θ* of a two-joint robot, the mass matrix is

M(\theta) = \left[

3*ba*12

\right],*M*(*θ*)=[3*b**a*12],

which has a determinant of 36-ab36−*ab* and eigenvalues \frac{1}{2} (15 \pm \sqrt{81 + 4 a b})21(15±81+4*ab*). What constraints must a*a* and b*b* satisfy for this to be a valid mass matrix? Select all that apply.

- a < 6
*a*<6 - b > 6
*b*>6 - a > b
*a*>*b* - a = b
*a*=*b* - a<b
*a*<*b* - a < \sqrt 6
*a*<6

Q5. An inexact model of the UR5 mass and kinematic properties is given below:

M_{01} = \left[

100001000010000.0891591

\right], M_{12} = \left[

00−10010010000.280.1358501

\right], M_{23} = \left[

1000010000100−0.11970.3951

\right], *M*01=⎣⎢⎢⎢⎡100001000010000.0891591⎦⎥⎥⎥⎤,*M*12=⎣⎢⎢⎢⎡00−10010010000.280.1358501⎦⎥⎥⎥⎤,*M*23=⎣⎢⎢⎢⎡1000010000100−0.11970.3951⎦⎥⎥⎥⎤,

M_{34} = \left[

00−1001001000000.142251

\right], M_{45} = \left[

10000100001000.09301

\right], M_{56} = \left[

100001000010000.094651

\right], *M*34=⎣⎢⎢⎢⎡00−1001001000000.142251⎦⎥⎥⎥⎤,*M*45=⎣⎢⎢⎢⎡10000100001000.09301⎦⎥⎥⎥⎤,*M*56=⎣⎢⎢⎢⎡100001000010000.094651⎦⎥⎥⎥⎤,

M_{67} = \left[

100000−10010000.082301

\right],*M*67=⎣⎢⎢⎢⎡100000−10010000.082301⎦⎥⎥⎥⎤,

G_1 = {\tt diag}([0.010267495893,0.010267495893, 0.00666,3.7,3.7,3.7]),*G*1=diag([0.010267495893,0.010267495893,0.00666,3.7,3.7,3.7]),

G_2 = {\tt diag}([0.22689067591,0.22689067591,0.0151074,8.393,8.393,8.393]),*G*2=diag([0.22689067591,0.22689067591,0.0151074,8.393,8.393,8.393]),

G_3 = {\tt diag}([0.049443313556,0.049443313556,0.004095,2.275,2.275,2.275]),*G*3=diag([0.049443313556,0.049443313556,0.004095,2.275,2.275,2.275]),

G_4 = {\tt diag} ([0.111172755531 ,0.111172755531 ,0.21942, 1.219, 1.219 ,1.219]),*G*4=diag([0.111172755531,0.111172755531,0.21942,1.219,1.219,1.219]),

G_5 = {\tt diag} ([0.111172755531 ,0.111172755531, 0.21942 ,1.219 ,1.219 ,1.219]),*G*5=diag([0.111172755531,0.111172755531,0.21942,1.219,1.219,1.219]),

G_6 = {\tt diag} ([0.0171364731454 ,0.0171364731454, 0.033822 ,0.1879 ,0.1879, 0.1879]),*G*6=diag([0.0171364731454,0.0171364731454,0.033822,0.1879,0.1879,0.1879]),

{\tt Slist} = \left[

001000010−0.08915900010−0.08915900.425010−0.08915900.8172500−1−0.109150.8172500100.00549100.81725

\right].Slist=⎣⎢⎢⎢⎢⎢⎢⎢⎡001000010−0.08915900010−0.08915900.425010−0.08915900.8172500−1−0.109150.8172500100.00549100.81725⎦⎥⎥⎥⎥⎥⎥⎥⎤.

Here are three versions for these UR5 parameters:

Given

*θ*=⎡⎣⎢⎢⎢⎢⎢⎢⎢⎢0*π*/6*π*/4*π*/3*π*/22*π*/3⎤⎦⎥⎥⎥⎥⎥⎥⎥⎥,*θ*˙=⎡⎣⎢⎢⎢⎢⎢⎢⎢0.20.20.20.20.20.2⎤⎦⎥⎥⎥⎥⎥⎥⎥,*θ*¨=⎡⎣⎢⎢⎢⎢⎢⎢⎢0.10.10.10.10.10.1⎤⎦⎥⎥⎥⎥⎥⎥⎥,g=⎡⎣00−9.81⎤⎦,Ftip=⎡⎣⎢⎢⎢⎢⎢⎢⎢0.10.10.10.10.10.1⎤⎦⎥⎥⎥⎥⎥⎥⎥,

use the function {\tt InverseDynamics}InverseDynamics in the given software to calculate the required joint forces/torques of the robot. The maximum allowable error for any number is 0.01, so give enough decimal places where necessary.

**Write the vector in the answer box and click “Run”:**

[1.11,2.22,3.33] for \left[

1.112.223.33

\right]⎣⎢⎡1.112.223.33⎦⎥⎤.

1

[0,0,0,0,0,0]

RunReset

### Week 02: Modern Robotics, Course 3: Robot Dynamics Coursera Quiz Answers

#### Quiz 01: Lecture Comprehension, Forward Dynamics of Open Chains (Chapter 8.5)

Q1. To derive the mass matrix M(\theta)*M*(*θ*) of an n*n*-joint open-chain robot, how many times would we need to invoke the recursive Newton-Euler inverse dynamics algorithm?

- 1 time.
**n***n*times.- It is not possible to derive the mass matrix from the Newton-Euler inverse dynamics.

Q2. When calculating the mass matrix M(\theta)*M*(*θ*) using the Newton-Euler inverse dynamics, which of these quantities must be set to zero? Select all that apply.

- \theta
*θ* **\dot{\theta}***θ*˙- \ddot{\theta}
*θ*¨ **The gravitational constant.****The end-effector wrench \mathcal{F}_{{\rm tip}}Ftip.**

#### Quiz 02: Lecture Comprehension, Dynamics in the Task Space (Chapter 8.6)

Q1. Converting the joint-space dynamics to the task-space dynamics requires an invertible Jacobian, as well as the relationships \mathcal{V} = J(\theta)\dot{\theta}V=*J*(*θ*)*θ*˙ and \dot{\mathcal{V}} = J\ddot{\theta} + \dot{J}\dot{\theta}V˙=*Jθ*¨+*J*˙*θ*˙, to find \Lambda(\theta)Λ(*θ*) and \eta(\theta,\mathcal{V})*η*(*θ*,V) in \mathcal{F} = \Lambda(\theta) \dot{\mathcal{V}} + \eta(\theta,\mathcal{V})F=Λ(*θ*)V˙+*η*(*θ*,V).

Why do you suppose we left the dependence on \theta*θ*, instead of writing it as a dependence on the end-effector configuration T \in SE(3)*T*∈*SE*(3), which would seem to be more aligned with our task-space view?

- Either T
*T*or \theta*θ*could be used; there is no reason to prefer one to the other. **The inverse kinematics of an open-chain robot does not necessarily have a unique solution, so we may not know the robot’s full configuration, and therefore the mass properties, given just T***T*.

#### Quiz 03: Lecture Comprehension, Constrained Dynamics (Chapter 8.7)

Q1. A serial-chain robot has n*n* links and actuated joints, but it is subject to k*k* independent Pfaffian velocity constraints of the form A(\theta)\dot{\theta}=0*A*(*θ*)*θ*˙=0. These constraints partition the n*n*-dimensional \tau*τ* space into orthogonal subspaces: a space of forces C*C* that do not create any forces against the constraints, and a space of forces B*B* that do not cause any motion of the robot. What is the dimension of each of these spaces?

**C***C*is (n-k)(*n*−*k*)-dimensional and B*B*is k*k*-dimensional.- C
*C*is n*n*-dimensional and B*B*is k*k*-dimensional. - C
*C*is k*k*-dimensional and B*B*is (n-k)(*n*−*k*)-dimensional. - C
*C*is k*k*-dimensional and B*B*is n*n*-dimensional.

Q2. Let the constrained dynamics of a robot be \tau = M(\theta)\ddot{\theta} + h(\theta,\dot{\theta}) + A^{\rm T}(\theta)\lambda*τ*=*M*(*θ*)*θ*¨+*h*(*θ*,*θ*˙)+*A*T(*θ*)*λ*, where \lambda \in \mathbb{R}^k*λ*∈R*k*. Let P(\theta)*P*(*θ*) be the matrix, as discussed in the video, that projects an arbitrary \tau \in \mathbb{R}^n*τ*∈R*n* to P(\theta)\tau \in C*P*(*θ*)*τ*∈*C*, where the space C*C* is the same C*C* from the previous question. Then what is P(\theta) \tau*P*(*θ*)*τ*? Select all that apply.

- M(\theta)\ddot{\theta} + h(\theta,\dot{\theta}) + A^{\rm T}(\theta)\lambda
*M*(*θ*)*θ*¨+*h*(*θ*,*θ*˙)+*A*T(*θ*)*λ* **P(\theta)(M(\theta)\ddot{\theta} + h(\theta,\dot{\theta}))***P*(*θ*)(*M*(*θ*)*θ*¨+*h*(*θ*,*θ*˙))**P(\theta)(M(\theta)\ddot{\theta} + h(\theta,\dot{\theta})) +A^{\rm T}(\theta)\lambda)***P*(*θ*)(*M*(*θ*)*θ*¨+*h*(*θ*,*θ*˙))+*A*T(*θ*)*λ*)

#### Quiz 04: Lecture Comprehension, Actuation, Gearing, and Friction (Chapter 8.9)

Q1. What is the typical reason for putting a gearhead on a motor for use in a robot?

**To increase torque (simultaneously reducing the maximum speed).**- To increase speed (simultaneously reducing the maximum torque).

Q2. Compared to a “direct drive” robot that is driven by motors without gearheads (G=1*G*=1), increasing the gear ratios has what effect on the robot’s dynamics? Select all that apply.

**The mass matrix M(\theta)***M*(*θ*) is increasingly dominated by the apparent inertias of the motors.- The mass matrix M(\theta)
*M*(*θ*) is increasingly dominated by off-diagonal terms. **The mass matrix M(\theta)***M*(*θ*) is increasingly dominated by constant terms that do not depend on the configuration \theta*θ*.- The robot is capable of higher speeds but lower accelerations.
**The significance of velocity-product (Coriolis and centripetal) terms diminishes.**

#### Quiz 05: Chapter 8.5-8.7 and 8.9, Dynamics of Open Chains

Q1. A robot system (UR5) is defined as

M_{01} = \left[

100001000010000.0891591

\right], M_{12} = \left[

00−10010010000.280.1358501

\right], M_{23} = \left[

1000010000100−0.11970.3951

\right], *M*01=⎣⎢⎢⎢⎡100001000010000.0891591⎦⎥⎥⎥⎤,*M*12=⎣⎢⎢⎢⎡00−10010010000.280.1358501⎦⎥⎥⎥⎤,*M*23=⎣⎢⎢⎢⎡1000010000100−0.11970.3951⎦⎥⎥⎥⎤,

M_{34} = \left[

00−1001001000000.142251

\right], M_{45} = \left[

10000100001000.09301

\right], M_{56} = \left[

100001000010000.094651

\right], *M*34=⎣⎢⎢⎢⎡00−1001001000000.142251⎦⎥⎥⎥⎤,*M*45=⎣⎢⎢⎢⎡10000100001000.09301⎦⎥⎥⎥⎤,*M*56=⎣⎢⎢⎢⎡100001000010000.094651⎦⎥⎥⎥⎤,

M_{67} = \left[

100000−10010000.082301

\right],*M*67=⎣⎢⎢⎢⎡100000−10010000.082301⎦⎥⎥⎥⎤,

G_1 = {\tt diag}([0.010267495893,0.010267495893, 0.00666,3.7,3.7,3.7]),*G*1=diag([0.010267495893,0.010267495893,0.00666,3.7,3.7,3.7]),

G_2 = {\tt diag}([0.22689067591,0.22689067591,0.0151074,8.393,8.393,8.393]),*G*2=diag([0.22689067591,0.22689067591,0.0151074,8.393,8.393,8.393]),

G_3 = {\tt diag}([0.049443313556,0.049443313556,0.004095,2.275,2.275,2.275]),*G*3=diag([0.049443313556,0.049443313556,0.004095,2.275,2.275,2.275]),

G_4 = {\tt diag} ([0.111172755531 ,0.111172755531 ,0.21942, 1.219, 1.219 ,1.219]),*G*4=diag([0.111172755531,0.111172755531,0.21942,1.219,1.219,1.219]),

G_5 = {\tt diag} ([0.111172755531 ,0.111172755531, 0.21942 ,1.219 ,1.219 ,1.219]),*G*5=diag([0.111172755531,0.111172755531,0.21942,1.219,1.219,1.219]),

G_6 = {\tt diag} ([0.0171364731454 ,0.0171364731454, 0.033822 ,0.1879 ,0.1879, 0.1879]),*G*6=diag([0.0171364731454,0.0171364731454,0.033822,0.1879,0.1879,0.1879]),

{\tt Slist} = \left[

001000010−0.08915900010−0.08915900.425010−0.08915900.8172500−1−0.109150.8172500100.00549100.81725

\right].Slist=⎣⎢⎢⎢⎢⎢⎢⎢⎡001000010−0.08915900010−0.08915900.425010−0.08915900.8172500−1−0.109150.8172500100.00549100.81725⎦⎥⎥⎥⎥⎥⎥⎥⎤.

Here are three versions for these UR5 parameters above:

Given

\theta = \left[

0*π*/6*π*/4*π*/3*π*/22*π*/3

\right]*θ*=⎣⎢⎢⎢⎢⎢⎢⎢⎡0*π*/6*π*/4*π*/3*π*/22*π*/3⎦⎥⎥⎥⎥⎥⎥⎥⎤,

\dot \theta = \left[

0.20.20.20.20.20.2

\right]*θ*˙=⎣⎢⎢⎢⎢⎢⎢⎢⎡0.20.20.20.20.20.2⎦⎥⎥⎥⎥⎥⎥⎥⎤,

\ddot{\theta} = \left[

0.10.10.10.10.10.1

\right]*θ*¨=⎣⎢⎢⎢⎢⎢⎢⎢⎡0.10.10.10.10.10.1⎦⎥⎥⎥⎥⎥⎥⎥⎤,

\mathfrak{g} = \left[

00−9.81

\right]g=⎣⎢⎡00−9.81⎦⎥⎤,

\mathcal{F}_{\text{tip}} = \left[

0.10.10.10.10.10.1

\right]Ftip=⎣⎢⎢⎢⎢⎢⎢⎢⎡0.10.10.10.10.10.1⎦⎥⎥⎥⎥⎥⎥⎥⎤

use the function {\tt MassMatrix}MassMatrix in the given software to calculate the numerical inertia matrix of the robot. The maximum allowable error for any number is 0.01, so give enough decimal places where necessary.

**Use Python syntax to express a matrix in the answer box:**

[[1.11,2.22,3.33],[4.44,5.55,6.66],[7.77,8.88,9.99]] for \left[

1.114.447.772.225.558.883.336.669.99

\right]⎣⎢⎡1.114.447.772.225.558.883.336.669.99⎦⎥⎤.

- 1
- [[0,0,0,0,0,0],[0,0,0,0,0,0],[0,0,0,0,0,0],[0,0,0,0,0,0],[0,0,0,0,0,0],[0,0,0,0,0,0]]

Q2. Referring back to Question 1, for the same robot system and condition, use the function {\tt VelQuadraticForces}VelQuadraticForces in the given software to calculate the Coriolis and centripetal terms in the robot’s dynamics. The maximum allowable error for any number is 0.01, so give enough decimal places where necessary.

**Use Python syntax to express a vector in the answer box:**

[1.11,2.22,3.33] for \left[

1.112.223.33

\right]⎣⎢⎡1.112.223.33⎦⎥⎤.

- 1
- [0,0,0,0,0,0]

Q3. Referring back to Question 1, for the same robot system and condition, use the function {\tt GravityForces}GravityForces in the given software to calculate the joint forces/torques required to overcome gravity. The maximum allowable error for any number is 0.01, so give enough decimal places where necessary.

**Use Python syntax to express a vector in the answer box:**

[1.11,2.22,3.33] for \left[

1.112.223.33

\right]⎣⎢⎡1.112.223.33⎦⎥⎤.

- 1
- [0,0,0,0,0,0]

Q4., Referring back to Question 1, for the same robot system and condition, use the function {\tt EndEffectorForces}EndEffectorForces in the given software to calculate the joint forces/torques required to generate the wrench \mathcal{F}_{{\rm tip}}Ftip. The maximum allowable error for any number is 0.01, so give enough decimal places where necessary.

**Use Python syntax to express a vector in the answer box:**

[1.11,2.22,3.33] for \left[

1.112.223.33

\right]⎣⎢⎡1.112.223.33⎦⎥⎤.

- 1
- [0,0,0,0,0,0]

Q5. Referring back to Question 1, for the same robot system and condition plus the known joint forces/torques

\tau = \left[

0.0128−41.1477−3.78090.03230.03700.1034

\right] *τ*=⎣⎢⎢⎢⎢⎢⎢⎢⎡0.0128−41.1477−3.78090.03230.03700.1034⎦⎥⎥⎥⎥⎥⎥⎥⎤,

use the function {\tt ForwardDynamics}ForwardDynamics in the given software to find the joint acceleration. The maximum allowable error for any number is 0.01, so give enough decimal places where necessary.

**Use Python syntax to express a vector in the answer box:**

[1.11,2.22,3.33] for \left[

1.112.223.33

\right]⎣⎢⎡1.112.223.33⎦⎥⎤.

- 1
- [0,0,0,0,0,0]

Q6. Assume that the inertia of a revolute motor’s rotor about its central axis is 0.005 kg m^22. The motor is attached to a zero-inertia 200:1 gearhead. If you grab the gearhead output and spin it by hand, what is the inertia you feel?

- 200 kg m^22
- 1 kg m^22
- 0.005 kg m^22

### Week 03: Modern Robotics, Course 3: Robot Dynamics Coursera Quiz Answers

#### Quiz 03: Lecture Comprehension, Point-to-Point Trajectories (Chapter 9 through 9.2, Part 1 of 2)

Q1. A point robot moving in a plane has a configuration represented by (x,y)(*x*,*y*). The path of the robot in the plane is (1+ 2\cos (\pi s), 2\sin(\pi s)), \; s \in [0,1](1+2cos(*πs*),2sin(*πs*)),*s*∈[0,1]. What does the path look like?

- An ellipse.
- A sine wave.
- A semi-circle.
- A circle.

Q2. Referring back to Question 1, assume the time-scaling of the motion along the path is s = 2t, \; t \in [0, 1/2]*s*=2*t*,*t*∈[0,1/2]. At time t*t*, where 0 \leq t \leq 0.50≤*t*≤0.5, what is the velocity of the robot (\dot{x},\dot{y})(*x*˙,*y*˙)?

- (-4 \pi \sin(2 \pi t), 4 \pi \cos(2 \pi t))(−4
*π*sin(2*πt*),4*π*cos(2*πt*)) - (-2\sin(2 \pi t), 2\cos(2 \pi t))(−2sin(2
*πt*),2cos(2*πt*))

Q3. True or false? For a trajectory \theta(s(t))*θ*(*s*(*t*)), the acceleration \ddot{\theta}*θ*¨ is \frac{d\theta}{ds}\ddot{s}*dsdθ**s*¨.

- True
- False

Q4. Let \mathcal{V}_sV*s* be the spatial twist that takes X_{s,{\rm start}}*Xs*,start to X_{s,{\rm end}}*Xs*,end in unit time. Which is an expression for the constant screw path that takes X_{s,{\rm start}}*Xs*,start (at s=0*s*=0) to X_{s,{\rm end}}*Xs*,end (at s=1*s*=1)?

- \exp([\mathcal{V}_s s]) X_{s,{\rm start}}, \; s \in [0,1]exp([V
*s**s*])*Xs*,start,*s*∈[0,1] - X_{s,{\rm start}}\exp([\mathcal{V}_s s]), \; s \in [0,1]
*Xs*,startexp([V*s**s*]),*s*∈[0,1]

#### Quiz 02: Lecture Comprehension, Point-to-Point Trajectories (Chapter 9 through 9.2, Part 2 of 2)

Q1. For a fifth-order polynomial time scaling s(t)*s*(*t*), t \in [0,T]*t*∈[0,*T*], what is the form of \ddot{s}(t)*s*¨(*t*)?

**Third-order polynomial**- Fourth-order polynomial
- Fifth-order polynomial

#### Quiz 03: Lecture Comprehension, Polynomial Via Point Trajectories (Chapter 9.3)

Q1. True or false? Third-order polynomial interpolation between via points ensures that the path remains inside the convex hull of the via points.

- True
**False**

Q2. A robot has 3 joints and it follows a motion interpolating 6 points: a start point, an end point, and 4 other via points. The interpolation is by cubic polynomials. How many total coefficients are there to describe the motion of the 3-DOF robot over the motion consisting of 5 segments?

**60**- 30

Q3. Referring again to Question 2, imagine we constrain the position and velocity of each DOF at the beginning and end of the trajectory, and at each of the 4 intermediate via points, we constrain the position (so the robot passes through the via points) but only constrain the velocity and acceleration to be continuous at each via point. Then how many total constraints are there on the coefficients describing the joint motions for all motion segments?

**60**- 30
- Quiz 04: Chapter 9 through 9.3, Trajectory Generation

Q1. Consider the elliptical path in the (x,y)(*x*,*y*)-plane shown below. The path starts at (0,0)(0,0) and proceeds clockwise to (1.5,1)(1.5,1), (3,0)(3,0), (1.5,-1)(1.5,−1), and back to (0,0)(0,0). Choose the appropriate function of s \in [0,1]*s*∈[0,1] to represent the path.

<image: https://d3c33hcgiwev3.cloudfront.net/imageAssetProxy.v1/KmuEUwnlEeiIOArs7r1YtA_37e2876750ce29ec1ffb0a4889fc703e_week3_elliptical-path.jpg?expiry=1665014400000&hmac=QBqjGjryxxevKjWBt_Afk3Y7EGCUWY8OztTlC_lA65w>

- x = 3 (1 – \cos 2 \pi s)
*x*=3(1−cos2*πs*) - y = \sin 2 \pi s
*y*=sin2*πs* - x = 1.5 (1 – \cos 2 \pi s)
*x*=1.5(1−cos2*πs*) - y = \sin 2 \pi s
*y*=sin2*πs* - x = 1.5 (1 – \cos s)
*x*=1.5(1−cos*s*) - y = \sin s
*y*=sin*s* - x = \cos 2 \pi s
*x*=cos2*πs*

y = 1.5 (1 – \sin 2 \pi s)*y*=1.5(1−sin2*πs*

Q2. Find the fifth-order polynomial time scaling that satisfies s(T) = 1*s*(*T*)=1 and s(0) = \dot{s}(0) = \ddot{s}(0) = \dot{s}(T) = \ddot{s}(T) = 0*s*(0)=*s*˙(0)=*s*¨(0)=*s*˙(*T*)=*s*¨(*T*)=0.

Your answer should be only a mathematical expression, a polynomial in t*t*, with coefficients involving T*T*. (Don’t bother to write “s(t) = *s*(*t*)=”, just give the right-hand side.

Q3. If you want to use a polynomial time scaling for point-to-point motion with zero initial and final velocity, acceleration, and jerk, what would be the minimum order of the polynomial

Q4. Choose the correct acceleration profile \ddot{s}(t)*s*¨(*t*) for an S-curve time scaling.

- A

<image: https://d3c33hcgiwev3.cloudfront.net/imageAssetProxy.v1/98PK-t7_EeeK2w4Lcly5FA_36dd58791fdeb9e5078c77589aef5616_ex05_1-01.jpg?expiry=1665014400000&hmac=8jz4kMHnZg18LLOkByxvPSVDm36AK7UChlf14OwUjyo>

- B

<image: https://d3c33hcgiwev3.cloudfront.net/imageAssetProxy.v1/S-eOht8AEeeK2w4Lcly5FA_eab5dfe0dfd05a5346dd0e1f3b9a0b30_ex05_2-01.jpg?expiry=1665014400000&hmac=Zam_yStud3bWx-JB7Car46Eq02lRbY8_tRJYEP687EU>

- C

<image: https://d3c33hcgiwev3.cloudfront.net/imageAssetProxy.v1/QnjatN8AEeeK2w4Lcly5FA_c72cb6462b524c6c55c870e1f930740e_ex05_3-01.jpg?expiry=1665014400000&hmac=0UdD5NhyVXP1i0JFguFsxApYEFC4ZN8AJTy2MXujQ7s>

- D

<image: https://d3c33hcgiwev3.cloudfront.net/imageAssetProxy.v1/M388Jd8AEeeVFRLN7DX_0g_fdec568a0c601a3f8292087790a225b2_ex05_4-01.jpg?expiry=1665014400000&hmac=KBBfiyXH7rQz_7goBVVAnRokX3i4x3DndQ0IWgW20u8>

Q5. Given a total travel time T = 5*T*=5 and the current time t = 3*t*=3, use the function {\tt QuinticTimeScaling}QuinticTimeScaling in the given software to calculate the current path parameter s*s*, with at least 2 decimal places, corresponding to a motion that begins and ends at zero velocity and acceleration

Q6. Use the function {\tt ScrewTrajectory}ScrewTrajectory in the given software to calculate a trajectory as a list of N=10*N*=10 SE(3)*SE*(3) matrices, where each matrix represents the configuration of the end-effector at an instant in time. The first matrix is

X_{{\rm start}} = \left [

1000010000100001

\right]*X*start=⎣⎢⎢⎢⎡1000010000100001⎦⎥⎥⎥⎤

and the 10th matrix is

X_{{\rm end}} = \left[

0100001010001231

\right].*X*end=⎣⎢⎢⎢⎡0100001010001231⎦⎥⎥⎥⎤.

The motion is along a constant screw axis and the duration is T_f = 10*Tf*=10. The parameter {\tt method}method equals 3 for a cubic time scaling. Give the **9th matrix (one before X_{{\rm end}} Xend) **in the returned trajectory. The maximum allowable error for any matrix entry is 0.01, so give enough decimal places where necessary.

**Use Python syntax to express a matrix in the answer box:**

[[1.11,2.22,3.33],[4.44,5.55,6.66],[7.77,8.88,9.99]] for \left[

1.114.447.772.225.558.883.336.669.99

\right]⎣⎢⎡1.114.447.772.225.558.883.336.669.99⎦⎥⎤.

- 1
- [[0,0,0,0],[0,0,0,0],[0,0,0,0],[0,0,0,1]]

Q7. Referring back to Question 6, use the function {\tt CartesianTrajectory}CartesianTrajectory in the MR library to calculate another trajectory as a list of N=10*N*=10 SE(3)*SE*(3) matrices. Besides the same X_{{\rm start}}*X*start, X_{{\rm end}}*X*end, T_f*Tf* and N = 10*N*=10, we now set {\tt method}method to 5 for a quintic time scaling. Give the **9th matrix (one before X_{{\rm end}} Xend) **in the returned trajectory. The maximum allowable error for any matrix entry is 0.01, so give enough decimal places where necessary.

**Use Python syntax to express a matrix in the answer box:**

[[1.11,2.22,3.33],[4.44,5.55,6.66],[7.77,8.88,9.99]] for \left[

1.114.447.772.225.558.883.336.669.99

\right]⎣⎢⎡1.114.447.772.225.558.883.336.669.99⎦⎥⎤.

- 1
- [[0,0,0,0],[0,0,0,0],[0,0,0,0],[0,0,0,1]]

RunReset

### Week 04: Modern Robotics, Course 3: Robot Dynamics Coursera Quiz Answers

#### Quiz 01: Lecture Comprehension, Time-Optimal Time Scaling (Chapter 9.4, Part 1 of 3)

Q1. When a robot travels along a specified path \theta(s)*θ*(*s*), torque or force limits at each actuator place bounds on the path acceleration \ddot{s}*s*¨. The constraints due to actuator i*i* can be written

\tau_i^{{\rm min}}(s,\dot{s}) \leq m_i(s) \ddot{s} + c_i(s)\dot{s}^2 + g_i(s) \leq \tau_i^{{\rm max}}(s,\dot{s})*τi*min(*s*,*s*˙)≤*mi*(*s*)*s*¨+*ci*(*s*)*s*˙2+*gi*(*s*)≤*τi*max(*s*,*s*˙).

What is one reason \tau_i^{{\rm min}}*τi*min and \tau_i^{{\rm max}}*τi*max might depend on \dot{s}*s*˙?

- The positive torque available from an electric motor typically increases as its positive velocity increases.
**The positive torque available from an electric motor typically decreases as its positive velocity increases.**

Q2. At a particular state along the path, (s,\dot{s})(*s*,*s*˙), the constraints on \ddot{s}*s*¨ due to the actuators at the three joints of a robot are: L_1 = -10, U_1 = 10*L*1=−10,*U*1=10; L_2 = 3, U_2 = 12*L*2=3,*U*2=12; and L_3 = -2, U_3 = 5*L*3=−2,*U*3=5. At (s,\dot{s})(*s*,*s*˙), what is the range of feasible accelerations \ddot{s}*s*¨?

**3 \leq \ddot{s} \leq 53≤***s*¨≤5- -10 \leq \ddot{s} \leq 12−10≤
*s*¨≤12

Q3. If the robot is at a state (s,\dot{s})(*s*,*s*˙) where no feasible acceleration \ddot{s}*s*¨ exists that satisfies the actuator force and torque bounds, what happens

- One or more of the actuators is damaged.
**The robot leaves the path.**- The robot must begin to decelerate, \ddot{s}<0
*s*¨<0. - Quiz 02: Lecture Comprehension, Time-Optimal Time Scaling (Chapter 9.4, Part 2 of 3)

Q1. Consider the figure below, showing 4 motion cones at different states in the (s,\dot{s})(*s*,*s*˙) space.

<image: https://d3c33hcgiwev3.cloudfront.net/imageAssetProxy.v1/3g3mIuT2Eee78hLOKMVkFA_f607c39ba0de00665594846ea21926f8_s-plane-cones-new.jpg?expiry=1665014400000&hmac=-k8-yJVL7PU8k0qkko-6NKkypE0umcNvKN7GnFe6hDo>

Which cone corresponds to U(s,\dot{s})=4, L(s,\dot{s})=-3*U*(*s*,*s*˙)=4,*L*(*s*,*s*˙)=−3?

- A
- B
- C
**D**

Q2. Considering the figure in Question 1, which cone corresponds to U(s,\dot{s})=4, L(s,\dot{s})=5*U*(*s*,*s*˙)=4,*L*(*s*,*s*˙)=5?

- A
- B
**C**- D

Q3. Which cone corresponds to U(s,\dot{s})=5, L(s,\dot{s})=2*U*(*s*,*s*˙)=5,*L*(*s*,*s*˙)=2?

**A**- B
- C
- D

Q4. Which cone corresponds to U(s,\dot{s})=-2, L(s,\dot{s})=-6*U*(*s*,*s*˙)=−2,*L*(*s*,*s*˙)=−6?

- A
**B**- C
- D

Q5. Assume a time scaling s(t) = \frac{1}{2}t^2*s*(*t*)=21*t*2. How is this time scaling written as \dot{s}(s)*s*˙(*s*)? (Note that this particular time scaling does not satisfy \dot{s}(1) = 0*s*˙(1)=0.)

**\dot{s} = \sqrt{2s}***s*˙=2*s*.- \dot{s} = \frac{1}{2}s^2
*s*˙=21*s*2.

#### Quiz 03: Chapter 9.4, Trajectory Generation

Q1. Four candidate trajectories (A, B, C, and D) are shown below in the (s,\dot{s})(*s*,*s*˙) plane. Select all of the trajectories that **cannot be correct**, regardless of the robot’s dynamics. **Note:** It is OK for the trajectory to begin and end with nonzero velocity. **(This is consistently one of the most incorrectly answered questions in this course, so think about it carefully!)**

<image: https://d3c33hcgiwev3.cloudfront.net/imageAssetProxy.v1/gbNGgeTbEeeRtwqRjGvJYg_87571e8df15ba4b3df7727ebbcf85300_s-plane-errors.jpg?expiry=1665014400000&hmac=m5SrNKwM4XCdIs1qH1xv_cHjkbzjYfdWfo_Nor6OOfk>

- A
- B
- C
- D

Q2. Four candidate motion cones at \dot{s} = 0*s*˙=0 (a, b, c, and d) in the (s,\dot{s})(*s*,*s*˙) plane are shown below. Which of these motion cones **cannot be correct** for any robot dynamics? (Do not assume that the robot can hold itself statically at the configuration.)

<image: https://d3c33hcgiwev3.cloudfront.net/imageAssetProxy.v1/CDMUTN6jEee5zAog3bNxIA_90ffe47d55d751049fc9a51a521c3f38_s-plane-errors_02-01.jpg?expiry=1665014400000&hmac=Iq7rlTaMghOFOYhsug5NPV0Ep4Oooe67kD8q1kuwGeQ>

- a
- b
- c
- d

### 3.

Question 3

We have been assuming forward motion on a path, \dot s > 0*s*˙>0. What if we allowed backward motion on a path, \dot s < 0*s*˙<0? This question involves motion cones in the (s, \dot s)(*s*,*s*˙)-plane when both positive and negative values of \dot s*s*˙ are available. Assume that the maximum acceleration is U(s, \dot s) = 1*U*(*s*,*s*˙)=1 (constant over the (s, \dot s)(*s*,*s*˙)-plane) and the maximum deceleration is L(s, \dot s) = -1*L*(*s*,*s*˙)=−1. For any constant s*s*, which of the following are the correct motion cones at the five points where \dot s*s*˙ takes the values \{-2, -1, 0, 1, 2\}{−2,−1,0,1,2}?

**1 point**

A

<image: https://d3c33hcgiwev3.cloudfront.net/imageAssetProxy.v1/lLTGZd6jEeedHw7X0zcEsg_6564b9361b064ed334c7f822ff25f265_ex03_1-01.jpg?expiry=1665014400000&hmac=z8gf3Wd9xhBtTDVqGe6Ba8kY0mIIs4WsO4J0msr-u3w>

B

<image: https://d3c33hcgiwev3.cloudfront.net/imageAssetProxy.v1/00NUK96jEee5zAog3bNxIA_14c9a76822602d2a763f8d6955e38887_ex03_2-01.jpg?expiry=1665014400000&hmac=WbzMSbuHORmaDHh5IIn0gipgjAth-PFG5YKbfLExQuE>

C

<image: https://d3c33hcgiwev3.cloudfront.net/imageAssetProxy.v1/579R4N6jEee8ZBJAMiIM0A_b7a3d7225eb02ee32507dac4d43f9924_ex03_3-01.jpg?expiry=1665014400000&hmac=rNYTqM7EivzCFvChNrmNM3zVvBdPpSZKwObc_Rgy5_c>

D

<image: https://d3c33hcgiwev3.cloudfront.net/imageAssetProxy.v1/F-aW996kEeedHw7X0zcEsg_a3b5ea7c6566d38e787c6982d1e7acb8_ex03_4-01.jpg?expiry=1665014400000&hmac=yyhFGBH4uspI9rsiAkUgFgk5tsodWxFA0WDRR1IAE3Y>

### 4.

Question 4

Referring back to Question 3, assume the motion starts at (s, \dot s) = (0, 0)(*s*,*s*˙)=(0,0) and follows the maximum acceleration U*U* for time t*t*. Then it follows the maximum deceleration L*L* for time 2t2*t*. Then it follows U*U* for time t*t*. Which of the following best represents the integral curve?

**1 point**

A

<image: https://d3c33hcgiwev3.cloudfront.net/imageAssetProxy.v1/VdJ88t6kEee8ZBJAMiIM0A_6bfbcadd6b7fbf218cf615fab8c1359d_ex04_1-01.jpg?expiry=1665014400000&hmac=p_Tp4UOxIcT31C3vOFVzq8uxQvdSFWWuhT34MqPvt9g>

B

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C

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D

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### 5.

Question 5

Below is a time-optimal time scaling \dot{s}(s)*s*˙(*s*) with three switches between the maximum and minimum acceleration allowed by the actuators. Also shown are example motion cones, which may or may not be correct.

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Without any more information about the dynamics, which motion cones must be incorrect (i.e., the motion cone is inconsistent with the optimal time scaling)? Select all that are incorrect (there may be more than one).

**1 point**

- A
- B
- C
- D
- E
- F
- G
- H

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