Simple problem about relative motion.

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A body with two spheres connected by a 10-meter rod is rolling down one incline and up another at 45 degrees. At a specific moment, the downward sphere has a velocity of 2 m/s and an acceleration of 3 m/s². The discussion focuses on calculating the upward sphere's velocity and the rod's angular velocity using the equation VB = VA + ω × rB/A. Participants express confusion over the calculations and the proper application of vector components and angular velocity. Clarification is sought on the equations used and the interpretation of terms like "ez."
ParrotPete
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Homework Statement


A body consisting of two spheres attached by a 10 meter solid long rod is rolling down a 45 degree incline at one end while rolling up a 45 degree incline on the other end.
At the instant when the connecting rod is horizontal the sphere rolling down has the velocity 2 m/s down the incline and the acceleration 3 m/s^2 down the incline. At this time, what is the velocity of the sphere rolling up and what is the angular velocity of the rod?

Homework Equations



VB = VA+ωXrB/A

The Attempt at a Solution


VA=2*(cos(45)*ei-sin(45)*ej
VB = VB*(cos(45)*ei+sin(45)*ej)
rA/B= 10*ei
ω=ω*ez
Doing the calculations
VB cos(45) = cos(45)
VB*sin(45) = -2*sin(45)+10ω

What am I doing wrong?
 
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ParrotPete said:
VA=2*(cos(45)*ei-sin(45)*ej
VB = VB*(cos(45)*ei+sin(45)*ej)
I guess that's VB = vB*(cos(45)*ei+sin(45)*ej), to avoid the pun.
rA/B= 10*ei
ek, perhaps?
ω=ω*ez
What's ez? Do you mean ei?
Doing the calculations
VB cos(45) = cos(45)
You've lost me - where did that come from?
 
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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