Problem statementSolution video |

**DISCUSSION THREAD**

Any questions??

Problem statementSolution video |

**DISCUSSION THREAD**

**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?

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Problem statementSolution video |

**DISCUSSION THREAD**

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**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:

**v _{A} = v_{C} + ω_{disk} x r_{A/C} =**

v

From these, you can solve for **ω _{disk}**

Applying the same procedure to acceleration:

**a _{A} = a_{C} + **

a

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

**a _{O} = a_{C} + α_{disk} x r_{O/C} - ω_{disk}^{2}**

Problem statementSolution video |

**DISCUSSION THREAD**

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**DISCUSSION and HINTS**

In this problem, end A of the bar is constrained to move along a straight horizontal path with a *constant* speed of *v _{A}*, whereas end B is constrained to move along a straight, angled path. As you can see in the animation below of the motion of the bar, the speed of B is NOT a constant (the acceleration of B is non-zero, and is, in fact, increasing as B moves along its path).

In your solution, it is recommended that you use the rigid body kinematics equations relating the motion of ends A and B:

**v**_{B} = **v**_{A} + **ω** x **r**_{B/A
}**a**_{B} = **a**_{A} + **α** x **r**_{B/A} - ω^{2}**r**_{B/A}

For these equations, you know: i) the magnitude and direction for the velocity of A; ii) that the acceleration of A is zero (constant speed along a straight path); and, iii) the direction for the velocity and acceleration of B. These two vector equations produce four scalar equations that can be solved for four scalar unknowns: v_{B}, a_{B}, *ω* and *α*.

** **

Problem statementSolution video |

**DISCUSSION THREAD**

**Discussion**

As you plan your solution, consider that for the velocity part of the problem, you will need to use a velocity equation for bar AB and two velocity equations for the disk:

- Write the velocity of point B referenced to point A on bar AB.
- Write the velocity of point O referenced to the contact point C on the disk. Note that since the disk does not slip, the velocity of point C is zero.
- Write the velocity of point B referenced to point O on the disk.

These equations together will provide you with the equations that you need for the angular velocities of the bar and disk.

Repeat this same process above for accelerations. Please note that the acceleration of C on the disk is NOT zero - instead, you know that its acceleration is strictly in the y-direction.

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Problem statementSolution video |

**DISCUSSION THREAD**

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*DISCUSSION*

This problem has two *no-slip* points on the disk (where below x is in the horizontal direction and y in the vertical direction):

- One at the bottom where the disk is in no-slip contact with the fixed ground (let's call this point C). For this point, we have
=**v**_{C}and**0***a*=_{Cx}*0*. This can be seen in the animation above in the position where B is directly BELOW O, and in no-slip contact with the ground. - One at the top of the disk where it is in contact with the translating block A (where we will refer to this as point D). For this point, we have
= v**v**_{D}_{A}and**i***a*=_{Dx}*a*. This can be seen in the animation above in the position where B is directly ABOVE O, and in no-slip contact with block A._{A}

For this problem you will need two sets of kinematics equations:

** v**_{A} = * v_{C}* +

Problem statementSolution video |

**DISCUSSION THREAD**

**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.

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Problem statementSolution video |

**DISCUSSION THREAD**

* 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} + omega x r_{B/A}**

a_{B} = a_{A} + α x r_{B/A} - ω^{2}*r_{B/A}

where * v_{B} = v_{B}*j, v_{A} = v_{A}*(cos(θ)*i + sin(θ)*j), a_{B} = a_{B}*j *and

*HINT*: The solution for this problem follows very closely that of * Example 2.A.7 *of the lecture book. If you get stuck on this problem, it is recommended that you review the solution video for Example 2.A.7 on the course website.

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?

Problem statementSolution video |

**DISCUSSION THREAD**

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* DISCUSSION*This is a standard relative motion type of question. You are given complete information on the velocity vectors for A and P (both

To do so, use the basic equation for relative velocity between two points: * v_{P/A} = v_{P} - v_{A}*. From this result, you are then asked to find the angle of the observed velocity of P, with this last step being just trig.

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* DISCUSSION*This is a standard cable/pulley kinematics problem. You need to write the length,

From the animation above, you can see that the speed of crate is never any greater than the speed of the worker, A. Can you see this result in your work?