# Questions on sum of angular speed and center of mass

I have been using MIT OpenCourseWare to teach myself Dynamics lately. The course I am studying is 2.003SC Engineering Dynamics. The lectures and learning resources provided by the course are really great, but I can say that I don't understand it 100%, so I think I should ask questions on this forum. I have asked a question related to the course and you might have seen it, and I believe in order to fully understand the course content I probably need to ask even more questions here in the near future.

In this thread, I am going to ask questions from problem set (pset) #2. Like other psets in the course, every problem is seperated into a conceptual question and a question involving calculation. I had done the calculation parts and got them correct; however, I am having trouble understanding the solutions for conceptual question #4 and #5. So, I am going to copy question #4 and #5 here and explain what I don't understand.

Conceptual Question #4 This question I got my answer right, but I have to admit I kind of guessed it. The question is :

Problem 4:

An amusement park ride: A ride at an amusement park consists of a rotating horizontal platform with an arm connected at point $$B$$ to the outside edge. The arm rotates at $$10 rev/min$$ with respect to the platform. The platform rotates at $$6 rev/min$$ with respect to the ground. The axes of rotation for both are in the $$z$$ direction, out of the plane of the page as shown in the drawing. Gravity acts in the $$z$$ direction. A seat is located at $$C$$ at the end of arm $$BC$$, which is $$2 m$$ long. The radius from $$A$$ to $$B$$ is $$6 m$$ long. The mass of the passenger is $$75kg$$. The reference frames defined in Figure 4 are as follows: $$O_{xyz}$$ is fixed to ground; $$A_{x_1 y_1 z_1}$$ is fixed to the platform at the axis of rotation. $$B_{x_2 y_2 z_2}$$ is fixed to the arm $$BC$$ at $$B$$ and rotates with the arm.

Concept question: What is the rotation rate experienced by the passenger with respect to the fixed coordinate system attached to the ground?

A. 4 rev/min

B. 10 rev/min

C. 16 rev/min

The answer is C. $$16 rev/min$$ which is the sum of angular speed of the arm with respect to the platform and the angular speed of the platform with respect to the ground. However, I don't think that the passenger doesn't go around point $$O$$ $$16$$ times per minute, so what is the meaning of this angular speed, why am I allowed to add two angular speeds of two different reference frames?

Conceptual Question #5

Problem 5:

Cart and rotating mass: A block of mass $$M_b$$ is constrained to horizontal motion by rollers. Let the acceleration of the block in the inertial frame $$O_{xyz}$$ be designated as $$\vec a_{A/O} = \ddot x \hat i$$ which is assumed to be known. Point A is fixed to the block. A massless rod $$AB$$ with length $$e$$ rotates about point $$A$$ with an angular rate of $$\dot \theta$$. A point mass, $$m$$, is attached to the end of the rotating rod at point $$B$$. This problem will appear several times over the span of this subject. It introduces one of the most important and most common problems in mechanical engineering—the unbalanced rotor.

Concept question: When considered as a two particle system, the block and the rotating mass have a center of mass which one could calculate as a function of the position of the rod as it rotates. When viewed from a fixed inertial frame external to the block, the center of mass of the system ...

A. Moves left when the ball is in the left half of the circular track

B. Moves right when the ball is in the left half of the circular track

C. Does not move left or right

My answer is A. Moves left when the ball is in the left half of the circular track because I think the center of mass keep rotating around the axis of rotation according to where the mass $$m$$ is. But the solution is C. Does not move left or right. The reason given in the solution is:

There are no external forces in the horizontal direction on the three body system, which consists of the block, the massless rod and the mass attached to the rod. Therefore the center of mass of the system in the horizontal direction does not move.

Now, this confuses me a lot. First, the block has a horizontal acceleration $$\vec a_{A/O} = \ddot x \hat i$$, so how come there is no external force acting on the system? Second, even if there is no external force acting on the system, I don't see why the center of mass $$G$$ stays in the same location since the location of center of mass is given by $$\vec r_{G/O} = \frac {\sum m_i \vec r_{i/O}}{\sum m_i}$$, so when the mass $$m$$ moves to the left of $$A$$, $$r_{m/O}$$ should be more to the left and the center of mass should average on the left of $$A$$. (I can't think of any equation or principle that relate external force to the center of mass's location, can anyone help me think of one?)

Lastly, the question and the solution say something about the term "unbalanced rotor" but I couldn't understand what is so unbalanced in this situation. Even after I have done the calculation part and got the right answer I couldn't really connect the result with the term (I couldn't picture what is happening in my head). Below are the calculation part questions and my solutions.

A. Find an expression for the total acceleration of the mass, $$m$$, in the fixed inertial frame $$O_{xyz}$$. That is find $$\vec a_{B/O}$$. Express the result in terms of $$x$$ and $$y$$ components in the $$O_{xyz}$$ frame. In other words use the unit vectors associated with the $$O_{xyz}$$ frame. Remember to include the contribution from $$\vec a_{A/O} = \ddot x \hat i$$ in your answer.

My solution:

I started with the kinematic equation

$$\vec a_{B/O} = \vec a_{A/O} + \vec a_{B/A} + 2\vec \omega_{A/O} \times \vec v_{B/A} + \vec \alpha_{A/O} \times \vec r_{B/A} + \vec \omega_{A/O} \times (\vec \omega_{A/O} \times \vec r_{B/A})$$

Which simplified to

$$\vec a_{B/O} = \ddot x \hat i + \dot \theta \hat k \times [\dot \theta \hat k \times (e \cos \theta \hat i + e \sin \theta \hat j)]$$

Thus,

$$\vec a_{B/O} = (\ddot x - e \dot \theta^2\cos \theta) \hat i -e \dot \theta^2\sin \theta \hat j$$

B. What is the magnitude and direction of the force that the rod applies to the mass, $$m$$? What is the force and direction that the rod places on the pivot point at $$A$$?

My solution:

Multiply the solution in part $$A$$ with mass $$m$$ to obtain the force.

$$\sum \vec F = m \vec a_{B/O}$$

$$\sum \vec F = m(\ddot x - e \dot \theta^2\cos \theta) \hat i -me \dot \theta^2\sin \theta \hat j$$

Since there is only a force $$\vec F_{rod}$$ exerted by the rod, then

$$\vec F_{rod} = m(\ddot x - e \dot \theta^2\cos \theta) \hat i -me \dot \theta^2\sin \theta \hat j$$

(This is also as far as the solution go, so I am going to ignore the expression for magnitude and angle)

The official solution continues on the result obtained in part B and explains that the force obtained in part B is equal and opposite to the force felt by the rotor's bearing. So the force exerted on rotor's bearing at point A is $$\vec F_A = \vec F_{rod} = -m\ddot x \hat i + e \dot \theta^2\cos \theta \hat i +me \dot \theta^2\sin \theta \hat j$$

From the expression of $$\vec F_A$$ what I understand is there is a force radially outward that keep pulling the rotor's bearing at point $$A$$ due to rotation of mass $$m$$ around $$A$$, is this what makes the rotor unstable? Moreover, the solution say that the unstable effect will perish if the axis of rotation pass through the center of mass, can anyone explain to me why is that so?

The full problem set and solution can be found here:

https://ocw.mit.edu/courses/mechanical-engineering/2-003sc-engineering-dynamics-fall-2011/velocity-acceleration-and-rotational-motion/