Converting angual velocity to linear velocity

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The discussion focuses on calculating various velocities and accelerations of an astronaut in a centrifuge with a radius of 10m and a given angular displacement equation. The angular velocity was determined to be 3 rad/s after calculating the angular displacement at t=5.0s. The user seeks guidance on converting angular velocity to linear velocity, considering the formula v = rω, and expresses confusion about unit conversions. It is clarified that radians can be effectively canceled out in the calculations since they are a ratio of arc length to radius. The conversation emphasizes the importance of understanding the relationships between angular and linear measurements in rotational motion.
crybllrd
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Homework Statement


An astronaut is being tested in a centrifuge. The centrifuge has a radius of 10m and, in starting, rotates \theta=0.30t^{2}, there t is in seconds and \theta in radians. When t=5.0s, what are the magnitudes of the astronaut's (a) angular velocity, (b) linear velocity, (c) tangential acceleration, and (d) radial acceleration?


Homework Equations



Kinematics equations

The Attempt at a Solution



For (a), I used \theta=.30(5)^{2}=7.5rad, plugging in 5.0s for t. Then knowing initial angular velocity = 0, time = 5s, and now \theta=7.5rad, I used a kinematics equation to determine final angular velocity to be 3rad/s.

Part (b) is where I'm stuck. I was thinking to convert 3 rad/s to degrees, then divide by 360 degrees to get 4.7 rotations a second, but not sure where to go from there (or if I'm on the right track).
 
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Well you can formulate the expression for v using v=rω and you know that ω=dθ/dt.
 
Thanks for the fast reply.
Where does that leave me in units? Seems like it would be (rad*m)/s. Is there a way I can divide out the radians?
 
crybllrd said:
Thanks for the fast reply.
Where does that leave me in units? Seems like it would be (rad*m)/s. Is there a way I can divide out the radians?

a radian is the ratio of arc length to radius so essentially, you can leave out the radian with no consequence.
 
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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