B Doubt regarding the terms used in the solution

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The terms Lcm and Icm refer to the angular momentum and moment of inertia of a sphere, respectively, as measured from the center of mass. The "cm" indeed stands for center of mass, indicating that these calculations are based on the sphere's center of mass axis. There is a discussion about a potential misprint in the solution regarding the translational velocity, suggesting it should be "v" instead of "vο." This clarification is important for accurate understanding and application of the concepts. Overall, the terms and their usage in the solution are confirmed to be correct, with the noted misprint being the only issue.
tbn032
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In the solution, the term Lcm and Icm is used. Explain the meaning of these terms? I think cm stands for centre of mass. why that is used in the subscript?does the term angular momentum from the centre of mass of the sphere makes sense? Is the term Lcm and Icm stand for angular momentum of the sphere from the axis passing through the centre of mass of the sphere and moment of inertia of sphere from the axis passing through the centre of mass of the sphere respectively.
 
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tbn032 said:
View attachment 313101 Is the term Lcm and Icm stand for angular momentum of the sphere from the axis passing through the centre of mass of the sphere and moment of inertia of sphere from the axis passing through the centre of mass of the sphere respectively.
Yes.
 
Also in the solution it is written "suppose the translational velocity of the sphere after it start rolling,is vο"is this a misprint ?, and it should have been written v instead of vο.
 
tbn032 said:
Also in the solution it is written "suppose the translational velocity of the sphere after it start rolling,is vο"is this a misprint ?, and it should have been written v instead of vο.
Yes!
 
For fun I was trying to use energy considerations to determine the depth to which a solid object will sink in a fluid to reach equilibrium. The first approach that I tried was just to consider the change in potential energy of the block and the fluid as the block is lowered some unknown distance d into the fluid similar to what is shown in the answer to this post. Upon taking the limit as the vessel's cross sectional area approaches infinity I have an extra factor of 2 in the equilibrium...