Force acting on the center of mass of a rolling disk

AI Thread Summary
A uniform disk is subjected to a force F at its highest point, leading to confusion about the relationship between angular and linear accelerations. The moment of inertia is calculated as MR^2/2, and the torque is determined to be FR, resulting in an angular acceleration of 2F/RM. The discussion highlights a critical point that the acceleration of the center of mass (c.o.m) is expressed as aR, but it is pointed out that the assumption of rolling without slipping is not stated. This raises questions about the validity of the derived equations and the conditions under which they apply. Clarification on the rolling condition is essential for accurate analysis of the forces and accelerations involved.
Fibo112
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Hello. The following situation I thought out confuses me so I am wondering where my mistake lies.
A uniform disk of mass M and radius R sits on its edge. A string is attached to the highest point and pulled with a Force F in the x direction.

The moment of inertia of the disk is MR^2/2 making the angular momentum about the center of mass MR^2w/2 where w is the angular velocity. The torque about the center of mass seems to be FR. Since torque is the derivative of angular momentum we have FR=MR^2a/2 where a is the angular acceleration.
This means the angular acceleration is equal to 2F/RM. The acceleration A of the c.o.m is equal to aR, so it is equal to 2F/M, making the force acting on the body equal to 2F/M*M= 2F?
 
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Fibo112 said:
This means the angular acceleration is equal to 2F/RM. The acceleration A of the c.o.m is equal to aR
What makes you say that the acceleration of the center of mass is equal to aR? Nobody said that the disk was rolling without slipping.
 

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