How to Determine Velocity at Gear Center Using Point C?

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To determine the velocity at the center of the gear using point C, the relationship between the velocities at points B and C is crucial. The angular velocity at point C was initially calculated incorrectly, leading to confusion about the correct velocity values. The equation v = ωr is applied to find velocities, but the relationship v_C = v_B cos(30) is suggested for clarity. A sketch is recommended to visualize the problem better and aid in calculations. Ultimately, understanding the connections between the points is essential for accurate results.
WhiteWolf98
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Homework Statement



f9f2b36b52142b4cee9169ed9f521b20.png


Homework Equations


##v=\omega r##

The Attempt at a Solution


So, using the equation, one can work out the velocity at point ##B##.
##v_B=\omega_{AB} \cdot r_B##
##v_B=6(0.4)=2.4~ ms^{-1}##I then tried working out the angular velocity at point ##C## using the instantaneous centre of zero velocity (I'm not sure of this is the correct next step. I don't think it is). But this ends up giving:
##\cos60= \frac {0.6} {r_{B/IC}}##
##r_{B/IC}=1.2##

Using ##v=\omega r##, this gives the angular velocity at point ##C## as ##2~rad/s##. This can't be right though since the velocity comes out as ##0.2## using ##r=0.1~m##

The answer is ##1.04~ms^{-1}##
 

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WhiteWolf98 said:
I then tried working out the angular velocity at point C using the instantaneous centre of zero velocity
Not sure what that means, or how you got the next equation.
Think about the velocity of C. What is the relationship between that and B's velocity?
 
Could it be that ##v_C={v_B} \cos30##?
 
WhiteWolf98 said:
Could it be that ##v_C={v_B} \cos30##?
Quite so.
 
How can I use the velocity at point ##C## to work out the velocity at the centre of the gear?
 
WhiteWolf98 said:
How can I use the velocity at point ##C## to work out the velocity at the centre of the gear?
When in doubt, make a quick sketch:
upload_2018-12-11_17-40-16.png
 

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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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