Rotational Motion Homework: K1+U1=K2+U2

AI Thread Summary
The discussion centers on the application of the conservation of energy equation K1 + U1 = K2 + U2 in a rotational motion problem involving a cylinder on a tabletop. The confusion arises regarding the presence of rotational kinetic energy, as the table is described as frictionless and the string is attached to the cylinder's center. However, it is clarified that the cylinder rolls without slipping, indicating that there is indeed friction acting at the point of contact, allowing for both translational and rotational motion. The key point is that the cylinder's rolling motion requires consideration of rotational energy despite the table's frictionless nature. Understanding the distinction between the frictionless table and the friction necessary for rolling is crucial for solving the problem correctly.
Muath Mushtaha
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Homework Statement


http://serv1.gulfup.cc/i/00003/ev7u8r04c18k.jpg

Homework Equations


K1+U1=K2+U2

The Attempt at a Solution


I have the solution i am just confused about one point,
in the solution, it considred both the translational and rotational kinetic energy for the cylinder on the tabletop,
but since the table is frictionless and the string is attached to its center, shouldn't it not rotate, thus only have translational energy and not rotational energy?
 
Last edited by a moderator:
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Muath Mushtaha said:

Homework Statement


http://serv1.gulfup.cc/i/00003/ev7u8r04c18k.jpg

Homework Equations


K1+U1=K2+U2

The Attempt at a Solution


I have the solution i am just confused about one point,
in the solution, it considred both the translational and rotational kinetic energy for the cylinder on the tabletop,
but since the table is frictionless and the string is attached to its center, shouldn't it not rotate, thus only have translational energy and not rotational energy?
The problem states that the cylinder rolls without slipping. That means that the table surface is not frictionless.
 
Last edited by a moderator:
It clearly states that the cylinder does not slip, essentially the distance traveled is only by rolling. So just consider the rotational motion.
 
Frictionless are mechanical parts.
 

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