Any questions??

Discussion and hints:

It is recommended that you use an observer attached to the boom. As we have discussed in class, your choice of observer directly affects four terms in the acceleration equation: ω and α  (how the observer moves), and the relative velocity and relative acceleration terms (what the observer sees). Note that the remainder of the discussion here is based on having the observer attached to the boom.

The boom shown above has TWO components of rotation:

• a rotation rate of Ω about a fixed axis (the “+” Y-axis), and,
• a rotation rate of θ_dot about a moving axis (the “+” z-axis)

(Be sure to make a clear distinction between the lower case and upper case symbols.)

Therefore, the angular velocity of the wheel is given by:

ω = Ω J + θ_dot k

The angular acceleration vector α is simply the time derivative of the angular velocity vector ω : α = dω/dt. In taking this time derivative,

• Recall that the J-axis is fixed. Since J is fixed, then dJ/dt = 0.
• Recall that the k-axis is NOT fixed. Knowing that, how do you find dk/dt?

With the observer attached to the boom, what motion does the observer see for point P? That is, what are (v_P/O)_rel and (a_P/O)_rel?

NOTE: Pay particular attention to the motion of the reference point O. What path does O follow? And, based on that, how do you write down the acceleration vector of point O, a_O?

Any questions??

Discussion and hints:

It is recommended that you use an observer attached to the wheel. As we have discussed in class, your choice of observer directly affects four terms in the acceleration equation: ω and α  (how the observer moves), and the relative velocity and relative acceleration terms (what the observer sees). Note that the remainder of the discussion here is based on having the observer attached to the wheel.

The wheel shown above has TWO components of rotation:

• a rotation rate of ω₁ about a fixed axis (the “+” Y-axis), and,
• a rotation rate of ω₂ about a moving axis (the “+” z-axis)

(Be sure to make a clear distinction between the lower case and upper case symbols.)

Therefore, the angular velocity of the wheel is given by:

ω = ω₁J + ω₂ k

The angular acceleration vector α is simply the time derivative of the angular velocity vector ω : α = dω/dt. In taking this time derivative,

• Recall that the J-axis is fixed. Since J is fixed, then dJ/dt = 0.
• Recall that the k-axis is NOT fixed. Knowing that, how do you find dk/dt?

With the observer attached to the wheel, what motion does the observer see for points A and B? That is, what are (v_A/O)_rel and (a_A/O)_rel, and (v_B/O)_rel and (a_B/O)_rel?

NOTE: Pay particular attention to the motion of the reference point O. What path does O follow? And, based on that, how do you write down the acceleration vector of O, a_O?

Any questions??

Discussion and hints:

The disk shown above has TWO components of rotation (note that ω₀ = 0):

• a rotation rate of θ_dot about a fixed axis (the “+” K-axis), and,
• a rotation rate of ω_disk about a moving axis (the “-” x-axis)

(Be sure to make a clear distinction between the lower case and upper case symbols.)

Therefore, the angular velocity of the disk is given by:

ω =θ_dot* K – ω_disk i

The angular acceleration vector α is simply the time derivative of the angular velocity vector ω : α = dω/dt. In taking this time derivative,

• Recall that the K-axis is fixed. Since K is fixed, then dK/dt = 0.
• Recall that the i-axis is NOT fixed. Knowing that, how do you find di/dt?

Any questions??

Discussion and hints:

It is RECOMMENDED that you choose to put your observer on the disk. In that case, the ω and α  that go into your acceleration equation will be that of the disk.

The discussion below follows a choice of putting the observer on the disk.

The disk shown above has TWO components of rotation:

• a rotation rate of ω₀ about a fixed axis (the “-” Z-axis), and,
• a rotation rate of ω_disk about a moving axis (the “+” y-axis)

(Be sure to make a clear distinction between the lower case and upper case symbols.)

Therefore, the angular velocity of the disk is given by:

ω = -ω₀K + ω_disk j

The angular acceleration vector α is simply the time derivative of the angular velocity vector ω : α = dω/dt. In taking this time derivative,

• Recall that the K-axis is fixed. Since K is fixed, then dK/dt = 0.
• Recall that the j-axis is NOT fixed. Knowing that, how do you find dj/dt?

Any questions??

Discussion and hints:

Let’s first take a look at the motion of point B. This motion of B is shown in the simulation results below.

The motion of B is circular, with the center of the path located at point D. This is expected since link BD is pinned to ground at point D.

Now, let’s attach an observer to link AC. Keep in mind that this observer is unaware that they are moving. The motion of B that this observer sees is straight, with this straight path aligned with the collar sliding on link AC. Since the xy-axes are attached to AC, this relative motion is in the x-direction:

(v_B/A)_rel = v_rel i = d_dot i

(a_B/A)_rel = a_rel i = d_ddot i

where d_dot and d_ddot are unknowns (you will solve for these in your analysis). You can see this is the following animation shown from the perspective of the observer attached to link AC.

Discussion

• Shown above left is the motion of the mechanism as seen by a fixed observer. Note that the stationary observer sees P moving on a circular path centered at point A.
• And, shown above right is the motion of P as seen by an observer attached to the slotted arm. Note that the moving observer sees P moving on a circular path centered at point C.

It is suggested that you employ the view of the observer on the slotted arm in your analysis.

Any questions?? Please ask/answer questions regarding this homework problem through the “Leave a Comment” link above.

Any questions??

Discussion:

Let’s first take a look at the motion of point D. This motion of D is shown in the simulation results below.

The motion of D is circular, with the center of the path located at point A. This is expected since link AD is pinned to ground at point A.

Now, let’s attach an observer to link BE. Keep in mind that this observer is unaware that they are moving. The motion that this observer sees is straight, with this straight path aligned with the slot cut into link BE. You can see this is the following animation shown from the perspective of the observer attached to link BE.

Any questions?? Please ask/answer questions regarding this homework problem through the “Leave a Comment” link above.

Discussion

An animation from the simulation of the motion of the robotic arm system is shown above for a fixed observer. The motion of end-effector B is quite complicated, as can be seen. However, if we attach an observer to section AD of the end link of the arm, the observed motion of B is greatly simplified. In fact, the observer simply sees a back-and-forth motion of B in the x-direction, as shown in the animation below.

HINTS: The velocity of point B can be found from the following moving reference frame velocity equation:

v_B = v_A + (v_B/A)_rel + ω × r_B/A

With the observer on link AD, we know the following:
v_A = (L θ_dot) i
ω  = φ_dot k
(v_B/A)_rel = b_dot i

Use similar logic for writing down the acceleration of B. Be careful in writing down the expression for a_A ; in particular, a_A ≠ 0. (A is traveling on a circular path with a constant speed.)

Any questions??

DISCUSSION
In some sense, this is a very standard kinematics of rigid bodies problem. A rigid link connects points A and B. To relate the motion of these two points, you will need the following kinematics equations:

v_A = v_B + ω x r_A/B 
a_A = a_B + α x r_A/B – ω² r_A/B

The nuance in this problem is the acceleration of point A. From what we learned earlier in the semester, the acceleration of a point can be written in terms of its path coordinates; that is, here the acceleration of A can be resolved into its tangential and normal components. As you proceed on this problem, you first need to recognize the tangential and normal unit vectors, e_t and e_n, for the motion of A. Then write down the acceleration of A, first in its path components, and then in its Cartesian component. In the end, you will have two scalar equations coming from the rigid body acceleration equation in terms of two unknowns.

Ask and answer questions here. Help others with thoughtful responses to their questions. You will learn from the experience.

DISCUSSION and HINTS
This mechanism is made up of three links: AB, BD and DE. You are given the rotation rate of link AB, and are asked to find the angular velocities and angular accelerations of links BD and DE. At the position shown, it is known that ω_AB = constant.

From the animation below, we are reminded that B and D move on circular arc paths with centers at A and E, respectively. The velocities of B and D are always perpendicular to the lines connecting the points back to the centers of the paths. Can you visualize the location of the instant center (IC) of link BD as you watch this animation? For the position of interest in this problem, you see in the animation that the velocities for all points on BD are the same – is this consistent with the location of the IC for BD at that position?

Velocity analysis
Where is the instant center of link BD? What does this location say about the angular velocity of BD?

Acceleration analysis
Write a rigid body acceleration equation for each of the three links in the mechanism:

a_B = a_A + α_AB x r_B/A – ω_AB²r_B/A
a_D = a_E + α_DE x r_D/E – ω_DE²r_D/E
a_B = a_D + α_BD x r_B/D – ω_BD²r_B/D

Combining together these three vector equations in a single vector equation will produce two scalar equations in terms of α_BD and α_DE.

Shown below is a freeze-frame of the motion of the mechanism at a time corresponding to the instant on which this question is based. As can be seen the velocity for all three points on link BD are the same (in both magnitude and direction) This is consistent with the conclusion that you would make based on knowing that the IC for link BD is at infinity for this instant where AB and DE are parallel.