Calculate Mass of Cylinder Given Tension & Distance

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SUMMARY

The discussion focuses on calculating the mass of a solid cylinder when a 300 g mass is suspended from a string wrapped around it. The mass falls 54 cm in 3.0 seconds, leading to a calculated tension of -2.9 N. Key equations include F = ma and τ = Iα, with the understanding that the radius of the cylinder cancels out in the calculations, allowing for the determination of the cylinder's mass without needing its radius.

PREREQUISITES
  • Understanding of Newton's second law (F = ma)
  • Familiarity with rotational dynamics (τ = Iα)
  • Knowledge of tension in strings and its effects on motion
  • Basic principles of kinematics for falling objects
NEXT STEPS
  • Study the relationship between tension and acceleration in systems involving pulleys
  • Learn about the moment of inertia for solid cylinders
  • Explore the concept of rolling motion and its equations
  • Investigate how to derive equations that involve both linear and angular motion
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Students in physics, particularly those studying mechanics, as well as educators looking for examples of rotational dynamics and tension in systems.

doub
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Homework Statement



A light string is wrapped around a solid cylinder and a 300 g mass hangs from the free end of the string, as shown. When released, the mass falls a distance 54 cm in 3.0 s. Calculate the mass of the cylinder?


Homework Equations



not sure


The Attempt at a Solution



I have gotten as far as to determine the Tension on the string of -2.9N. Without known the radius of the cylinder I feel that the mass cannot be determined.

Granted these are study question but if the final is going to be like this...yikes
 
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hi doub! :smile:
doub said:
A light string is wrapped around a solid cylinder and a 300 g mass hangs from the free end of the string, as shown. When released, the mass falls a distance 54 cm in 3.0 s. Calculate the mass of the cylinder?

Without known the radius of the cylinder I feel that the mass cannot be determined.

the motion depends on what i call the rolling mass of the cylinder, I/r2, which is always half the actual mass! :wink:

carry on … you should find that the radius cancels out :smile:
 
The best I can get is

\alpha = (T1 + T2)/I

= (Fr - Tr)/1/2 mr2

= (mra -(ma+mg)r)/ 1/2 mr2

don't see how the radiii cancel each other though
 
doub said:
\alpha = (T1 + T2)/I

= (Fr - Tr)/1/2 mr2

= (mra -(ma+mg)r)/ 1/2 mr2

sorry, i don't understand any of this :redface:

you should have an F = ma equation and a τ = Iα equation
 

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