# Homework H2.I - Fa22

NOTE: At time marker 13:58 on the solution video for this problem, a sign error was made in writing down the y-component of the position vector rE/D. Below shows the needed subsequent corrections in the solution due to this sign error.

Before starting this problem, make note of the type of motion for each component in the mechanism:

• Links OA and BC are in pure rotation about their centers of rotation O and C, respectively. From this, we know that the paths of points A, B and C are circular, as seen in the animation below.
• Block E is in pure translation.
• Links AB and DE have both translational and rotation components of motion.

Question: What are the locations of the instant centers (ICs) of AB and DE at this instant? Reflect back on the observations above in answering this. What do these locations say about the angular velocities of AB and DE at this position?

HINTS:
Once you have found the angular velocities for all of the links, you can then tackle the acceleration analysis.

• For finding the angular acceleration of links AB and BC, use the following rigid body acceleration equations:
aA = aO + αOA x rA/O - ωOA2rA/O
aB = aC + αBC x rB/C - ωBC2rB/C
aB = aA + αAB x rB/A - ωAB2rB/A
This will give you the equations that you need to solve for the desired angular accelerations.
• Repeat the above for link DE to determine its angular acceleration:
aD = aC + αBC x rD/C - ωBC2rD/C
aE = aD + αDE x rE/O - ωDE2rE/O

# Homework H2.J - Fa22

Any questions??

Before starting this problem, make note of the type of motion for each component in the mechanism:

• Block B is in pure translation, and moves to the left with a constant speed of vB.
• The disk is pinned to ground at its center C and moves in pure rotation; that is, C is the center of rotation of the disk. Because of this, the velocity of A is in the vertical direction.
• Slider D is in pure translation, and, at the instant of interest, D moves in the same direction as does Block B.
• Link AB has both translational and rotation components of motion.

Question: Where is the location of the instant center (IC) of AD at this instant? Reflect back on the observations above in answering this. You can use the location of the IC for AB to either find the angular velocity of AD, or to check your answer found from vector analysis.

HINTS:

• Use the rigid body kinematics equations relating the motion of the contact point on the disk (call that point E) to the center C of the disk to find the angular velocity and angular acceleration of the disk:
vE = vC + ωdisk x rE/C
aE = aC + αdisk x rE/C - ωdisk2 rE/C
• Use the rigid body kinematics equations relating the motion of points A and B to find the angular velocity and angular acceleration of the link AD:
vD = vA + ωAD x rD/A

Shown below is a freeze-frame capture of the animation of the motion for this mechanism at the position of interest (when link AC is horizontal). Note that the velocity of A is to the RIGHT at this position. The direction of motion quickly changes to the LEFT at the point where D becomes the IC of AD, as it should.

# Homework H2.G - Fa22

Discussion

The animation above shows the motion of the mechanism over a range of input angles of link OA. For a given position, envision the location of the instant centers (ICs) for links AB and CD. Do the directions and magnitudes for the velocities of points B, C and D agree with the location of these ICs?

Shown below if a freeze-frame of the mechanism motion at the position for which you are asked to do analysis. From this figure, where are the two ICs for links CD and BC? In particular, how does the position of the IC for AB relate to the relative sizes of the speeds of points A, C and B? What is the angular velocity of link AB at this position?

Any questions??

# Homework H2.F - Fa22

Discussion and hints

From the simulation results above, we see that point A travels on a cycloidal path, with the velocity vector for A being tangent to this path, as expected. In addition, the acceleration of A points inward to the path (again, expected).  The angle between v and a is initially obtuse, implying that A is initially decreasing in speed. At some point, this angle because acute indicating that the speed of A begins to increase. In fact, this rate of change of speed becomes very large as A approaches the surface on which the disk rolls. Do you know why?

The velocity analysis is a straight-forward application of our rigid body kinematics equations where we write a velocity equation for each rotating member:

vA = vC + ωdisk x rA/C =ωdisk x rA/C
vA = vB + ω
AB x rA/B = vBi + ωAB x rA/B

From these, you can solve for ωdisk and ωAB.

Applying the same procedure to acceleration:

aA = aCαdisk x rA/C - ωdisk2 rA/C
aA = aB + α
AB x rA/B - ωABrA/B

produces too few equations for the number of unknowns. It is recommended that you also use the following equation in your acceleration solution:

aO = aC + αdisk x rO/C - ωdisk2 rO/C

# Homework H2.C - Fa22

Discussion and hints:

The solution for the velocity and acceleration of end B is a straight-forward application of the rigid body velocity and acceleration equations for member AB:

vB = vA + omega x rB/A
aB = aA + α x rB/A - ω2*rB/A

where vB = vBj_hat, aB = aB*j_hat and aA = aA*(cos(θ)*i_hat + sin(θ)*j_hat)Each of the two vector equations above represents two scalar equations, providing us with the necessary equations to solve for vB, ω, aB and α. All of the observations made above can be predicted by the above kinematics equations. Instant centers (later on in the course) can prove useful in providing explanations.

For the inclination angle used in the above simulation, we see that point B moves DOWNWARD along the vertical wall as A moves up along the incline. As B moves onto the same horizontal plane as A, the acceleration of B becomes very large (although A continues to move with a constant speed). Can you provide a physical explanation for this?

If we now consider a steeper inclination angle for A, as used above, we see that end B initially moves UPWARD along the wall; however, at some point B reverses its direction and begins to move DOWNWARD along the wall. Can you provide a physical explanation for this? Note also that the acceleration of B becomes very large as B moves onto the same horizontal plane as A, as it was for the initial value of inclination angle.

What is the value of the incline angle theta that defines the boundary between the types of initial motions for bar AB shown in the above two simulations? For the numerical value of the angle theta provided in the problem statement, which of the two simulations above agree with your results?

# Homework H2.D - Fa22

Animation of motion

Note that as link BD passes through the horizontal orientation, the velocity of B (as well as the angular velocity of AB) instantaneously goes to zero, yet the acceleration of B is not zero. In a couple days when we cover instant centers of zero velocity we will be able to predict when point B has zero velocity.

Solution plan

1. Note that since D is traveling on a straight line with constant speed, the acceleration of D is zero. Therefore, we can write: a_B = a_D + alpha_BD x r_B/D - omega_BD^2*r_B/D = alpha_BD x r_B/D - omega_BD^2*r_B/D.
2. Note that since A is a pin joint, the acceleration of A is zero. Therefore, we can write: a_B = a_A + alpha_AB x r_B/A - omega_AB^2*r_B/A = alpha_AB x r_B/A - omega_AB^2*r_B/A.
3. Equate the expressions of a_B  from 1. and 2. above. This gives you two SCALAR equations in order to solve for two unknowns, alpha_AB and alpha_BD.

Note that you need to solve the velocity problem first. Follow the above procedure to do so. Here. the velocity of D is known as v_D = -v_D*j_hat.

# Homework H2.A - Fa22

Discussion

Note that the plate rotates about point O. Therefore, O is the center of the circular paths of points A and B. From the animation above, we see that the velocities of A and B are tangent to their circular paths, as expected. The accelerations of A and B are NOT perpendicular to the paths of A and B since the speeds of A and B are increasing in time (and consequently, A and B each have positive tangential components of acceleration).

Initially, the acceleration for these two points is nearly aligned with velocity, since the speeds are small and therefore the centripetal components of acceleration are small. Near the end of the first revolution of the plate, the speeds have increased to the point where the centripetal components of acceleration dominate, and acceleration is nearly perpendicular to the path.

Solution hints
For Part a) of this problem, it is recommended that you use the rigid body kinematics equations using point O as the reference point, since the velocity and acceleration of O are zero. That is, you should use v_B = v_O + Ω x r_B/O and a_B = a_O + Ω_dot x r_B/O - Ω^2*r_B/O. Repeat the process for finding the velocity and acceleration of A.

For Part b) of this problem, it is recommended that you use the rigid body kinematics equations with point A first. This will give you the equations needed to find Ω_dot. Then, use the rigid body kinematics equations to find the acceleration of B.