30 thoughts on “Homework H5.E - Sp24”

1. Should the frictional torque be shown in the FBD?

   1. Yes.

2. In which direction is the frictional moment acting on the system?

   1. I could be incorrect, but I think that in order to determine which way the pully will act, you compare the moment caused by mass A and mass B. Whichever is bigger will cause the wheel to turn in the direction of the moment it causes. The direction of the friction will be opposite to this motion.

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      1. I am still confused how would you isolate the moment just caused by A or B without finding the tensions? Or am I just overcomplicating it?

         1. How I understood it was that because the system is released from rest, we can assume initial tension T = mg for both sides. Therefore, you can use the masses to determine the moment about the center O. Whichever of those is greater will be the initial direction of motion (and alpha) with the frictional torque opposing this direction.

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            1. Tension is not equal to the weight of the block when released from rest. At rest, there is an external force preventing motion, but when that external force is removed and the block is no longer being forced to remain at rest, there will be a net acceleration of each block. This can't occur if the tension force on each block is equal to the block's weight. This problem can be solved by using the kinematics of the circular drum as well as force-balance equations.

3. Is this in the horizontal or vertical plane?

   1. Motion is in the vertical plane. The gravitational forces create the motion.

   2. You can see it is in the vertical plane in the way the motion of the pulley works. I like to look at it in the way that only if gravity is the main force other than tension you can see that it would move only vertically.

4. Once the direction of the frictional moment is determined, I found that the solving process for this problem is very similar to 5.A.11 in the lecturebook.

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5. Once we find the direction of the frictional torque, do we have to incorporate it into one of our equations or does it just affect a certain parameter's direction?

   1. It will be incorporated into the Euler equation for the pulley because it causes a moment.

6. One of the major steps in this problem is to determine the accelerations of both blocks A and B, preferably in terms of alpha. You will want to use a rigid body equation, however determining the vector between a point with known acceleration (point o, for instance) and the centers of A or B is impossible. If you are stuck on this, how can the accelerations of A and B be modelled on the drum, which has known dimensions?

7. Would the moment of inertia for the pully be that of two disks (one for each radius) added together?

   1. The moment of inertia would just be mk0^2. They give you the radius of gyration in the problem which is what you use for your r in mr^2=I. The mass is the mass of the pulley.

8. Never mind, I remembered it gives you k

9. How should the radius of gyration K_o be used in this problem?

   1. Consider the formula for the radius of gyration and what components make it up. The formula for radius of gyration is ko = sqrt( I / m ), so from this formula we can algebraically find the mass moment of inertia for the drum.

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10. Can anyone explain why it is best practice here to use Newton equations on the blocks, and not on the pulley? Does it make the solution process easier? We have the same amount of unknowns in either case.

    1. Keeping in mind that the tensions in the cables are not equal to the weights of the blocks, completing Newton's equations would include reaction forces at O, introducing new unknowns that don't easily relate to angular acceleration. This method would still leave Ta and Tb unknown, making it more difficult than considering the newton equations on the blocks. The relationships between the accelerations of blocks A and B with the accelerations of points C and E on the pulley are vital for solving this problem.

11. How do I find the relationship between the applied torque, moment of inertia, and angular acceleration for the pulley?

12. Note that the frictional torque of the pulley is given in N*m. This means that we can incorporate friction into the moment equation without finding where on the pulley it acts. All we actually need is the direction the pulley will move in, which should then give us the sign of the frictional torque.

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13. If we don't know the values of two two tension forces, how can we find the direction of rotation?

    1. You have to find the accelerations at points C and E and relate them to the tension forces in A and B using the sum of forces equations. From there you can use those tensions in the moment equation and get an idea of which way the disk is spinning.

    2. As mentioned by Hayden, we can assume an initial tension of T=mg for both sides as the system is released from rest. Using the masses and the different radii r, we can calculate the torque about point O for blocks A and B using Tau = rF. The direction of angular acceleration will be towards the block with the greater torque, with the frictional torque opposing this direction.

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14. When doing your free body diagrams make sure to take into account the spin of the pulley. This will change the direction of the two blocks with one being the opposite of the other.

15. When writing the acceleration equation, would it be correct to say the angular velocity term is zero since the system is released from rest?

    1. yes

16. It might be useful to remember the equation a = angular acceleration*R. Also, it might be useful to draw FBD for each of the masses to get the directions right, and one for the pulley as a whole.

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