Force acting on the center of mass of a rolling disk

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SUMMARY

The discussion centers on the dynamics of a uniform disk subjected to a force F applied at its highest point. The moment of inertia is established as MR²/2, leading to a calculated angular acceleration of 2F/RM. However, a critical point raised questions the assumption that the acceleration of the center of mass (c.o.m) is equal to aR, as the disk's rolling condition is not specified. This highlights the importance of clarifying whether the disk rolls without slipping to accurately determine the relationship between linear and angular accelerations.

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  • Understanding of Newton's laws of motion
  • Familiarity with rotational dynamics and moment of inertia
  • Knowledge of torque and its relationship to angular momentum
  • Concept of rolling motion and conditions for rolling without slipping
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  • Study the principles of rolling motion and the conditions for rolling without slipping
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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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