Linear speed and rotational quantity

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To determine the linear speed of a hula hoop with a total kinetic energy of 0.15 J, both translational and rotational kinetic energy must be considered. The relationship between linear speed and rotational speed is defined by the condition of rolling without slipping, which indicates that the linear speed is equal to the radius multiplied by the angular velocity. The total kinetic energy combines both translational (1/2 mv^2) and rotational (1/2 Iω^2) components. Understanding these relationships and applying the appropriate formulas will yield the required linear speed. The discussion emphasizes the need for clarity in linking linear and rotational dynamics in rolling objects.
baylorbelle
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What linear speed must a 6.0×10−2 hula hoop have if its total kinetic energy is to be 0.15 J ? Assume the hoop rolls on the ground without slipping.


So, i know that the formula for linear momentum is p=mv. however, the hoop is circular, meaning it has to have a rotational velocity (omega). I can't seem to figure out how the Joules play into any equation that helps link linear and rotational speeds. Anyone know of any formulas I can use for this problem?
 
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If something is both translating and rotating, how do you determine its total kinetic energy?

What does "rolling without slipping" tell you about how translational speed relates to rotational speed?
 
if something is rolling without slipping, it has rotational speed but not translational speed?
 
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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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