Chapter 4 Homework

Homework H4.R – Sp 26

17 Comments
Problem statement
Solution video

DISCUSSION THREAD

Discussion and hints
Recall the following four-step plan outline in the lecture book and discussed in lecture:

Step 1: FBDs
Draw a free body diagram of P.

Step 2: Kinetics (angular impulse/momentum and work/energy)
Note that all forces acting on P in the plane of the table point toward  the fixed point O. What does this say about the angular momentum of P about point O? Also, consider the work/energy equation for P.

Step 3: Kinematics
The kinematics of P are best written in terms of polar coordinates R and φ.

Step 4: Solve
Solve for the R and φ components of velocity of P from these equations.

Any questions?

17 thoughts on “Homework H4.R – Sp 26”

    1. That would not be a good assumption to make.

      If you draw an FBD of P, you will see that the only force acting on P in the plane of the tabletop is the cable force. This cable force is directed toward point O. As a consequence of this, the moment acting on the particle about O is zero. This means that the angular momentum of P about O is a constant. This does NOT mean the same thing as saying that the angular velocity of the cable is constant.

      Does this help?

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    1. Recall that angular momentum about a fixed point O is defined as:
      H_O = m*(r_P/O x v_P)
      = m*(R*e_R) x (R_dot*e_R + R*theta_dot*e_theta)
      = m*R^2*theta_dot*k
      For a problem like this where there is no moment about point O, we can say that H_O = constant. With that, you can see that the e_theta component of velocity is not constant. Instead, R^2*theta_dot is a constant.

      1. Would we be able to say that R^2*theta_dot is the same for R1 and R2 to set them equal to each other and solve for theta dot two? Or is this only applicable for the second part of the question to find R dot

        1. What you are proposing is that R^2*theta_dot = CONSTANT. This is a consequence of saying that angular momentum is conserved. For this problem, angular momentum is conserved, so your statement is correct here.

  3. It is helpful to consider the following: when solving for H2 and H1, at the instant the system is released from rest for H1, there is no velocity in the e_r direction. therefore the velocity gathered after finding a way to equate H1 and H2 will be in the e_phi direction (this is phi_dot). this means another equation is needed for solving for the velocity in the e_r direction (R_dot).

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    2. Consider using the work/energy equation along with conservation of angular momentum. The work done on the system will be that of the applied for F on the end of the cable with the work being equal to F times the amount of cable that is pulled through the hole in the table.

    1. This is a case of where you do not need to deal with phi_ddot. phi_ddot is part of the phi component of acceleration of P. Since this problem deals with only velocity through angular momentum and energy, phi_ddot does not come into play with your equations.

      To address the question of whether it is zero or not, it is not zero. If you make that assumption, you are assuming that phi_dot = constant. This would be in contradiction to what conservation of angular momentum says: R^2*phi_dot = constant.

  5. After doing my calculations I found that R_dot is positive. However, this doesn’t align with what is shown in the problem and I was wondering if it was because of the square root.

  7. Using work energy to find R double but I’m not sure what R1 double dot would be, would it be zero since at R=1.5 there is no velocity in the ehat_theta direction?

    1. I’m not sure of your question here. You do not need to deal with R_ddot in this problem. Here we are looking to find velocity components.

      Are you actually asking about R_dot? If so, then you are correct. At the initial state R_dot = 0 since the velocity has only a theta component.

      Let me know if I have not addressed your question.

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