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Discussion and hints

In order to create the animated GIFs shown below, a problem slightly different that that described in the homework statement has been solved. Here, the rotation rate of omega_OB is not constant, but instead, link OB is oscillating back and forth.

What you should sees from these animations is that the velocity of P as seen by a stationary observer is NOT tangent to the circular guide, whereas the velocity as seen by the observer on OB is tangent to the circular guide. Also, the observed acceleration of P always points to the inside of the circular guide, whereas the acceleration seen by a fixed observer does not always do so.

We need to treat this problem just as we have done for all planar mechanism kinematics in this course. For velocity analysis:

1. Write down a rigid body kinematics equation relating points A and P. This vector equation will have one unknown: omega_AP.
2. Write down a moving reference frame (MRF) equation kinematics equation relating points P and O. For this equation, it is recommended that you attach an observer onto link OB. This vector equation will have one unknown: v_rel (the velocity of P as seen by the observer on OB). Note that v_rel will be tangent to the circular guide.
3. Set the above two velocity equations in terms of v_P equal, and balance coefficients of the xy unit vectors. This will give you two equations in terms of the two unknowns listed above.

For acceleration analysis, repeat the above steps using the rigid body acceleration and MRF acceleration equations. Note that for the MRF equation, the acceleration of P as seen by the observer will have TWO components: one tangent to the circular guide (the observed rate of change of speed, vdot_rel) and one perpendicular to the circular guide (the observed centripetal component of acceleration, v_rel^2/R). In the end, you will have two equations in terms of vdot_rel and alpha_AP.

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Discussion

The motion of this four-bar mechanism is quite complicated. However, we can make some sense out of the velocities of points B and D by using instant centers (ICs). For a given position of the mechanism, can you visualize the location of the IC for link BD? And from this location, do the directions for the velocities for points B and D make sense?

HINTS:
You are asked to perform the complete kinematical analysis for the mechanism for the position shown below.

Velocity
For the velocity analysis, you are encouraged to use the IC center approach. Where is the IC for link BD located? What does the location of this IC say about the angular velocity of link BD? Use that result to determine the angular velocity of link DE.

Acceleration
This is a standard kinematics of a mechanism problem. For this, consider the following steps:

1. Write down the acceleration of point B using link AB.
2. Write down the acceleration expression for D using link DE.
3. Write down the acceleration expression for B using link DB and your result for the acceleration of D from Step 2.
4. Equate the expressions for the acceleration of B in Steps 1 and 3 above. From this, solve for the angular accelerations of links BD and DE.

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Discussion

Consider the motion of the system shown below,

along with a freeze-frame of this motion that follows:

Points B and D on the disk roll without slipping on the horizontal guide A and the wedge E, respectively. Because of this:

• B has a velocity v_B which points straight down, with the same speed v_A as the horizontal guide A.
• D has a velocity v_D which points directly to the left with the same speed as the wedge E.

The instant center (IC) of the disk (call it point C) therefore exists at the intersection of the perpendiculars to v_B and v_D. Note that point B is much closer to the IC than point D. What does this say about the relative sizes of v_D and v_B? Is this consistent with the speeds of D and B shown in the above animation?

HINTS:

• Use the above discussion points to locate the IC for the disk.
• Determine the angular speed of the disk based on the speed of B and the distance from B to C.
• The speed of the wedge E is found by using the angular speed from above and the distance from C to D.

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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 is 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?

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DISCUSSION

Note that with the disk not slipping on the outer cable, and with section BC of that cable being at rest, the velocity of point C on the disk has zero velocity. This means that the disk rolls without slipping on the cable, which is the same is it were rolling without slipping on a fixed vertical wall.

With the above in mind, consider the following:

• Let M represent the point on the left side of the pulley from which the outer cable leaves contact with the disk. Note that M is moving vertically upward with a speed of v_A.
• Write a rigid body kinematics equation relating the motion of points C and M: v_M = v_C + ω x r_M/C . That equation will produce an expression for ω, the angular speed of the disk.
• Then, you can use v_N = v_C + ω x r_N/C and  v_H = v_C + ω x r_H/C to find the velocities of point N and H on the disk. Since the inner cable does not slip on the disk, then v_N = v_D and v_H = v_Q.
• Also,  v_E = v_C + ω x r_E/C.

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DISCUSSION
For the velocity analysis of this mechanism, it is recommended that you write down three vector velocity equations: 1) relating the motion of O and C; 2) the motion of B and C; and, 3) relating the motion of A and B. Solve the resulting scalar equations from these vector equations for the angular velocities of AB and of the disk.

For the acceleration analysis, repeat the above process, but now use the three acceleration equations. Solve the resulting scalar equations from these vector equations for the angular accelerations of AB and of the disk. Please note that the acceleration of the no-slip contact point is NOT zero. Please read this page from the lecture book for an explanation as to why C can have a non-zero component of acceleration in the direction normal to the surface on which it rolls..

For example, the velocity and acceleration equations for link AB are given by:
v_B = v_A + ω_AB x r_B/A
a_B = a_A + α_AB x r_B/A - ω_AB²*r_B/A

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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:

v_B = v_A + ω x r_B/A
a_B = a_A + α x r_B/A - ω²*r_B/A

where v_B = v_B* j, a_B = a_B*j and a_A = a_A*(cos(θ)*i + sin(θ)*j). Each of the two vector equations above represents two scalar equations, providing us with the necessary equations to solve for v_B, ω, a_B 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?

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.

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