Rotational Dynamics Practice Problems

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The discussion revolves around two rotational dynamics problems involving pulleys and weights. In the first problem, a 2.00kg textbook and a 3.00kg hanging book are analyzed to find the tension in the cord and the moment of inertia of the pulley, with the user struggling to determine the mass of the pulley needed for calculations. The second problem involves a 50.0N frictionless pulley with weights of 75.0N and 125N, where the user is confused about calculating the force exerted by the hook on the ceiling, mistakenly summing the weights instead of considering the dynamics of the system. Suggestions highlight that the first problem can be simplified by directly calculating acceleration, while the second requires recognizing that the masses are in motion, affecting the forces involved. Clarifications on these concepts are essential for solving both problems effectively.
rwx1606
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Homework Statement


I'm stuck on two problems so I'll just write them both here.
A. A 2.00kg textbook rests on a frictionless, horizontal surface. A cord attached to the book passes over a pulley whose diameter is 0.150m, to a hanging book with mass 3.00kg. The system is released from rest, and the books are observed to move 1.20m in .800s. What is the tension in each part of the cord. What is the moment of inertia of the pulley about its rotational axis.
B.Two weights are connected by a very light flexible cord that passes over a 50.0N frictionless pulley of radius .300m. The pulley is a solid uniform disk and is supported by a hook connected to the ceiling. What force does the hook exert on the hook. The weight on the left is 75.0N, and the weight on the right is 125N.

Homework Equations


A.τ = Iα, I=1/2MR^2,\alpha=a/R

B.I=1/2MR^2

The Attempt at a Solution


A. For this problem I used Newton's 2nd Law for the two books to calculate the tension. The horizontal tension, which I'll call T1, was equal to 2a. The vertical tension, T2, is a mg+ma=29.4+3a. I then proceeded to calculate the net torque and got a (T1-T2)R. And set that equal to the moment of inertia I, times the angular acceleration. I converted the angular acceleration into terms of translational acceleration. So (T1-T2)R=1/2MR^2(a/R). After eliminate some variables, I get that the a=2(T1-T2)/M. The trouble is, I have three sets of equations to solve for the tensions, and acceleration but how do I get M, the mass of the pulley?
B.I don't get the concept in this question. I attempted to just add the weight of the pulley and the two weights and this was equal to 250N. I know the ceiling will pull up on the hook with the same force the pulley pulls down on the hook with. But the answer is 239N and I don't know how they got that. I know I'm missing something key, but can't quite figure it out.

Any help is appreciated! Thank you for taking the time to read.
 
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no one? :(
 
For the first question, you were given enough information in the question to find 'a' without having to solve for it with your system of equations. Figure out that, and things should become easier.

For the second question, I think the masses are supposed to be moving. So how would that change things?
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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