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DISCUSSION and HINTS
This mechanism is made up of three links: AB, BC and the wheel. You are given the constant rotation rate of link AB, and are asked to find the angular velocities and angular accelerations of link BC and of the wheel.

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

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_C = a_D + α_wheel x r_C/D - ω_wheel²r_C/D
a_B = a_C + α_BC x r_B/C - ω_BC²r_B/C

where D is the point on the wheel in contact with the ground. Note that the horizontal component of a_D is zero, whereas the vertical component of a_D is not zero. Combining together these three vector equations in a single vector equation will produce two scalar equations in terms of α_wheel and α_BC .

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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.

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DISCUSSION and HINTS

As discussed in lecture, we can locate the instant center (IC) for a rigid body through knowledge of the direction of motion for two points on that body. Here, we want to know the location of the IC for link BD. To this end:

• The path of B is circular with its center at A. Therefore, the velocity of B v_B is vertical.
• The path of D is circular with its center at E. Therefore, the velocity of D v_D is horizontal.
• Draw a line that is perpendicular to v_B , and draw a line that is perpendicular to v_D. The intersection of these two perpendiculars is the IC for link BD. See the animation below that shows how the location of the IC for for BD moves as the mechanism moves. It is called the "instant" center because its location changes from instant to instant.

Recall, also, that the speed of any point on a rigid body is equal to the angular speed of the body times the distance from that point to the IC for that body. This provides you with the equations needed to find the angular velocities of links BD and DE.

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DISCUSSION

As you watch the animation below, you see that C moves back and forth along a straight path. In general, C has a non-constant speed; therefore, with the path of C being straight, its acceleration is always directed along the path of C. Point B, on the other hand, moves along a circular arc path with a non-constant speed. Owing to the curvature of the path of B, the general acceleration of B is directed neither purely tangent nor purely normal to the path of B. Using our earlier understanding of the path description of motion, the normal component of acceleration of B is due to the centripetal term of v_B²/r. Note in the animation that for positions of B for which v_B = 0, the acceleration of B is tangent to the path.

HINTS
Velocity analysis
As recommended in lecture, we should write down a rigid body velocity equation for each rotating link in the mechanism. For this problem, we should use:

v_B = v_A + ω_AB x r_B/A
v_B = v_C + ω_BC x r_B/C

For these two equations, we know both v_A(= 0) and v_C . Equating these two equations gives two scalar equations in terms of ω_AB and ω_BC .

Acceleration analysis
Similarly for acceleration, we should use:

a_B = a_A + α_AB x r_B/A - ω_AB²r_B/A
a_B = a_C + α_BC x r_B/C - ω_BC²r_B/C

For these two equations, we know both a_A(= 0) and a_C . Equating these two equations gives two scalar equations in terms of α_AB and α_BC .

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DISCUSSION THREAD

At first read through, the motion of the disk rolling on the translating wedge looks complicated. It turns out not be be so complicated; rather, it can be described through a straight-forward expression of the rigid body kinematics equation for the disk. If we focus our attention on the disk by itself, we know the direction of of motion for two points on the disk: point O moving in the negative y-direction with an unknown speed of v_O, and since point C does not slip on the wedge, C moves in the negative x-direction with a known speed of v_C = v_A. This is as shown in the following figure.

From this, we can write the following vector kinematics equation for the disk:
v_C = v_O + ω x r_C/O

Although this is a single vector equation, it is made up of two scalar equations due to its x- and y-components. These two scalar equations are in terms of two unknowns: ω and v_O. For Part (a), solve these two equations.

For Part (b), you can now write the rigid body kinematics equations relating points O and B in order to find the vector velocity expression for point B:
v_B = v_O + ω x r_B/O.

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DISCUSSION and HINTS
Block A moves to the right with a speed of v_A, with this speed changing at a rate of a_A. The disk rolls without slipping on the block in such a way that C has a known speed of v_C, with this speed changing at a rate of a_C.

It is recommended that for this problem, you use the following rigid body kinematics equations relating the motion of the center of the disk C and the point on the disk that is in contact with block A (let's call that point E):

v_C = v_E + ω x r_C/E
a_C = a_E + α x r_C/E - ω²r_C/E

What do we know about the motion of point E? Point E has the same x-component of velocity and acceleration as does block A. The y-component of velocity of E is zero, but the y-component of acceleration of E is NOT zero. (Do you know why?)

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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 - ω²r_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.

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DISCUSSION and HINTS
The solution for this problem is a straight-forward application of our two rigid body kinematics equations relating the motion of point A back to point O:

v_A = v_O + ω x r_A/O
a_A = a_O + α x r_A/O - ω²r_A/O

where point O is fixed (zero velocity and zero acceleration).

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DISCUSSION and HINTS

The motion of point C is quite complicated. However, its path is defined by the combination of motion from the translation of A with the rotation of the plate. As expected, the velocity of C is always tangent to its path.

The solution for this problem is a straight-forward application of our rigid body kinematics equations relating the motion of point C back to point A:

v_C = v_A + ω x r_C/A

where point A is translating to the right (positive x-direction) and the plate is rotating CCW at a rate of ω.