We encourage you to interact with your colleagues here in conversations about this homework problem.

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Discussion and hints:

Note that since the top contact point of the disk rolls without slipping on the upper fixed surface, this contact point has zero velocity. In addition, the horizontal component of acceleration of that point is also zero (the vertical component of acceleration is NOT zero, however).

For the velocity problem, write down the velocity kinematics equation for the disk (where “C” is the contact point):

v_O = v_C + ω x r_O/C = ω x r_O/C

and solve for the angular velocity of the disk. For the acceleration problem, write down the acceleration kinematics equation for the disk:

a_O = a_C + α x r_O/C – ω²r_O/C = a_C j + α x r_O/C – ω²r_O/C

and solve for the angular acceleration of the disk.

Then, write down and use the acceleration rigid body kinematics equation relating points O and B.

Carefully study the velocity (BLUE) and acceleration (RED) information for the point on the circumference of the disk shown in the animation below. Recall that the velocity of the point on the disk in contact with the fixed upper surface is zero, and its acceleration has only a vertical component. Do you see this in the animation?

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Ask your questions here. And, consider answering questions of your colleagues here. Either way, you can learn.

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.

Discussion and hints:

The solution for this problem requires only a straight-forward application of the rigid body acceleration equation, relating the known accelerations of points A and C:

a_C = a_A + α x r_C/A – ω²r_C/A

Since this vector equation has two components (x and y), you have two scalar equations from which you can solve for ω and α.

Any questions??

We encourage you to interact with your colleagues here in conversations about this homework problem.

Discussion and hints:

Enlarged view

This problem is  a straight-forward application of the planar rigid body kinematics equations. The velocity and acceleration of the center of the disk are known, as well as the angular velocity and angular acceleration of this disk. To find the velocity and acceleration of point A on the disk, you can use the following:

v_A = v_O + omega x r_A/O
a_A = a_O + alpha x r_A/O – omega^2*r_A/O

Using the enlarged view above, we can see the interplay between velocity (blue) and acceleration (red) of point A. Using what we learned back in Chapter 1, can you identify times when the speed of A is increasing and when it is decreasing? (Hint: Look at the angle between the velocity and acceleration vectors.) Using the top view, we can see the path of point A – is that the path that you expected for A?

Discussion and hints:

This is a typical cable-pulley problem for us in this course. Here, we need to first write the length of the cable, L, in terms of the position variables of s_A and s_B, along with some constants. Since it is assumed that the cable does not break, go slack or stretch, the length L is constant. Take a time derivative of the expression above for L. This relates the speeds of blocks A and B.

Carefully study the animation of the motion of this cable-pulley system provided below. One of the questions asked of you is the speed of B when s_A = 0. What do you see in the animation below as s_A passes through zero? Does your equation above tell you the same thing?

Any questions??

Discussion and hints:

An observer riding along on block A will describe the motion of P relative to themself in terms of the radial distance r and the cable angle θ. Because of this, it is convenient to use the relative motion kinematics equations to write down the velocity and acceleration of particle P:  v_P = v_A + v_P/A and a_P = a_A + a_P/A , where the relative velocity and relative acceleration terms are written in polar form:

v_P/A = r_dot e_r + r*theta_dot e_θ

a_P/A = (r_ddot -r*theta_dot²) e_r + (r*theta_ddot + 2r_dot*theta_dot) e_θ

You will need to write the polar unit vectors e_r and e_θ in terms of the Cartesian unit vectors, i and j, at some point before you can add together terms.

QUESTION: What is the relationship between x_A and r? (Recall that the cable remains taut at all times.)

An animation showing the path of point P corresponding to θ_dot = constant is provided below. How does this true path differ from the view of the path as seen by the observer on A? Can you visualize that? HINT: The path seen by the observer on A is an inward spiral centered on the observer (point A).

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