Calculating Mass of Cylinder in Rotational Dynamics Problem

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SUMMARY

The discussion focuses on calculating the mass of a solid cylinder in a rotational dynamics problem involving a block of mass 2.0 kg hanging from a massless string. The block falls 82 cm in 2.0 seconds, leading to an acceleration of 0.41 m/s² and a final velocity of 0.82 m/s. The tension in the string is calculated to be 20.42 N. The participants confirm that the tangential acceleration of the cylinder's surface matches the block's acceleration due to the non-slipping condition of the string.

PREREQUISITES
  • Understanding of Newton's laws of motion
  • Familiarity with rotational dynamics concepts
  • Knowledge of free body diagrams
  • Basic equations of motion in physics
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  • Study the relationship between linear and angular acceleration in rotational systems
  • Learn how to derive mass from tension and acceleration in pulley systems
  • Explore the equations of motion for rotating bodies
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This discussion is beneficial for physics students, educators, and anyone interested in understanding the principles of rotational dynamics and the calculations involved in mass and acceleration in mechanical systems.

Kyleman
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A massless string is wrapped around a solid cylinder. A block of mass m=2.0 kg hangs from the string. When released, the block falls a distance 82 cm in 2.0 s. Calculate the mass of the cylinder.

Okay, for this problem, I started off with drawing free body diagrams for the block and the cylinder. I calculated the acceleration for the block to be .41 m/s^2, the final velocity of the block to be .82m/s and the force of tension of the string to be 20.42 N. Now, I think I need to calcuate the radius for the cylinder. Can I use the equation a=v^2/r to do that or is that the acceleration that is going towards the center of the cylinder?

Thanks for the help.
 
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I guess I'm assuming that the acceleration of the mass is the same acceleration as the cylinder, is that correct?
 
A point in/on the cylinder has two acceleration components - tangential and radial. The tangential acceleration of a point on the surface of the cylinder will be the same as the acceleration of the falling block since the string and the block (and therefore the outer surface of the block - the tangential acceleration) share the same linear displacement in time. since the string is in contact with the cylinder (not slipping on it) outer surface of the cylinder will travel at the same speed as the string and block at all times.
 

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