Understanding Velocity=r(omega)??

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Velocity is defined as the product of the radius and angular velocity, expressed as v = r(omega). Angular velocity is measured in radians per second, and when multiplied by the radius, it yields linear velocity. The relationship arises from the definition of arc length, where s = θr, and the connection between angular displacement and time. This relationship simplifies the understanding of motion in circular paths, as radians are defined based on the radius of a circle. Understanding these definitions clarifies why the equation holds true in circular motion.
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Homework Statement


I do not understand why velocity is equal to the radius times the angular velocity. Angular velocity is given in radians per second. How does this equal velocity when multiplied by the radius?

Thanks yall! I'm new here by the way.


Homework Equations


v=r(omega)


The Attempt at a Solution




??
 
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S=θr

where S is the Arc Length.

So think of it like r*θ/t
 
johnps said:

Homework Statement


I do not understand why velocity is equal to the radius times the angular velocity. Angular velocity is given in radians per second. How does this equal velocity when multiplied by the radius?

Thanks yall! I'm new here by the way.


Homework Equations


v=r(omega)


The Attempt at a Solution




??

What is the equation for the diameter of a circle, in terms of the radius? That should help it to make more sense.

Welcome to the PF, BTW.
 
By definition, s=θr. By definition, w=dθ/dt. That's why ds/dt=r*dθ/dt=r*w=v. Why the definition s=θr, you ask? That's how the radian was defined. It's simple, elegant, and leaves out a pesky constants that clutter up equations.
 
Aha, so s=(theta)(radius) because of the equation circumference = 2 (pi) (r) correct?

Sorry I'm not sure how to get the symbols yet. I'll figure it out.
 
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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