What is the Moment of Inertia of a Cylinder with a Mass Hanging from a Cord?

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Homework Help Overview

The discussion revolves around calculating the moment of inertia of a cylinder with a mass hanging from a cord. The problem involves concepts from rotational dynamics and kinematics, specifically relating to torque, angular acceleration, and linear acceleration.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants explore various methods to calculate the moment of inertia, including using torque equations and energy considerations. There are discussions about the use of average velocity and its impact on the calculations. Some participants question the assumptions made regarding initial and final velocities.

Discussion Status

The conversation is ongoing, with participants providing feedback on each other's calculations and questioning the validity of their approaches. There is no clear consensus yet, as multiple interpretations and calculations are being explored.

Contextual Notes

Participants are working under the constraints of a homework problem, which may limit the information available for their calculations. There are discussions about the correctness of values used for velocity and acceleration, indicating potential misunderstandings or miscalculations.

John O' Meara
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A cylinder of radius 20cm is mounted on an axle coincident with its axis so as to be free to rotate. A cord is wound on it and a 50-g mass is hung from it. If after being released, the mass drops 100cm in 12s, find the moment of inertia of the cylinder? Take down as positive.
Torque=I*(alpha), i.e., b*T=I*(alpha) => T=I*(alpha)/b ...(i)
m*g - T=m*a ...(ii), on adding i and ii, we get
I=m*b(g-a)/(alpha); now alpha=a/b, where a=acceleration, b=radius, T=tension, I=moment of inertia and s=1m.
v_avg=(v+u)/2=.04166 m/s: v^2=u^2+2*a*s therefore a=-u^2/2/s=-8.681*10^-4. Hence I=225.78 kg.m^2.
I also did the calculation from the energy point of view and got the same answer. Yet the answer in the book is 1.4kg.m^2. Can anyone show me where I went off track? Thanks very much.
 
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v_avg=(v+u)/2=.04166 m/s: v^2=u^2+2*a*s therefore a=-u^2/2/s=-8.681*10^-4. Hence I=225.78 kg.m^2.

Looks like you used the average velocity for the final velocity in determining 'a.'
 
Noein said:
Looks like you used the average velocity for the final velocity in determining 'a.'
Yes. And that calculated average velocity is wrong by a factor of 2, so the v used is off by a factor of 4.
 
I used the initial value for the velocity v, and now I get I=56.46kg.m^2; which is still the wrong answer. So I wounder what else is wrong? Any help would be welcome.
 
John O' Meara said:
I used the initial value for the velocity v, and now I get I=56.46kg.m^2; which is still the wrong answer. So I wounder what else is wrong? Any help would be welcome.
What initial value would that be? The initial velocity is zero. The final velocity at the end of 12 seconds can be calculated from the average and initial velocities, and the average velocity can be calculated from the distance and the time. Show us the calculation you did using what you believe to be the correct velocity. Then we can find your mistake.
 
The torque=I*(alpha); b*T=I*(alpha) therefore T=I*(alpha)/b
m*g - T = m*a therefore I=m*b^2(g/a-1) where (alpha)=a/b; v_avg=1/12 = 8.333*10^-2; therefore v=2*v_avg=16.666*10^-2m/s;
a=16.666*10^-2/12s=1.388*10^-2m/s/s.
I=.04*.05(9.8/1.388*10^-2 - 1) = 1.41kg.m^2. Much thanks for your help OlderDan.
 

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