Solving Belt Around Pulleys Homework: 0.6 m/s, 1.44 m/s2, 1.56 m/s2

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The discussion revolves around solving a physics homework problem involving a belt over two pulleys with given parameters. The normal acceleration at point A is 0.9 m/s², and the angular acceleration of pulley O1 is 3.6 rad/s². The belt velocity is calculated to be 0.6 m/s, with the belt acceleration determined to be 1.44 m/s². The total acceleration at point B is found to be 1.56 m/s² using the relevant equations for tangential and normal acceleration. The solution emphasizes the importance of free body diagrams and applying Newton's second law to derive necessary equations.
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Homework Statement


A belt passes over two pulleys O1 (radius R1= 0.4 m) and O2 (radius R2 = 0.6 m) as shown, and points A and B are on the rim of the pulleys. The normal acceleration at A is 0.9 m/s2 and the angular acceleration of pulley O1 is 3.6 rad/s2. Find (a) the belt velocity υb, (b) the belt acceleration ab, and (c) the acceleration of B.


Homework Equations


a n= v^2/R
angular acceleration and velocity equations

The Attempt at a Solution


I got a by using a n=v^2/R... so .9 = v^2/.4, v = .6 which is correct. I do not know how to go about finding the other values at all.

answers: (a) 0.6 m/s (b) 1.44 m/s2 (c) 1.56 m/s2
 
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Draw free body diagrams for your two pulleys, label all the forces acting on them, including torques, and use Newtons second law to derive equations that can help you solve for your unknowns. You may have to look up the moment of inertia of a pulley (a cylinder I suppose?)
 
How's VTech? haha, I just finished it, and I'll post now:

so we have:

r1=.4 m
r2=.6 m
Ana= .9 m/s^2 (this is normal acceleration for point A)
alpha1= 3.6 rad/s^2 (this is angular acceleration for O1)

In order to find Belt Velocity we...

Ana=(wa^2)(r1) => wa=sqrt((Ana)/r1)) = 1.5 => v=wr=(1.5)(.4)=.6m/s

In order to find Belt Acceleration we...

at (tangential acceleration) = (alpha1)(r1) = 1.44 m/s^2

In order to find acceleration of B:

we know that at above =1.44m/s^2

we can solve for Anb=(v^2)/(r2)=(.6^2)/(.6)=.6
so TOTAL acceleration of aT

aT=sqrt((at^2)+(Anb^2))=1.56 m/s^2

Sorry about the sloppy notation :-/
 
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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