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What is the net torque on the system

  • Thread starter Guruu
  • Start date
1. The problem statement, all variables and given/known data

A 4.60 kg counterweight is attached to a light cord, which is wound around a spool. The spool is a uniform solid cylinder of radius 7.00 cm and mass 1.00 kg.

(a) What is the net torque on the system about the point O?
Magnitude: 3.16 N·m


(b) When the counterweight has a speed v, the pulley has an angular speed ω = v/R. Determine the total angular momentum of the system about O.

____ v kg·m2/s

(c) Using the fact that torque = dL/dt and your result from (b), calculate the acceleration of the counterweight.
______ m/s2



2. Relevant equations

L=Iw ?
I=1/2mr^2 ?



3. The attempt at a solution

I've already done part A. My question involves part B. My best try was to calculate the angular momentum of the block and apply that to the whole system but that obviously doesn't work. My problem is with factoring in the counterweight when finding the angular momentum of hte whole system. I hate this unit, I'm pretty lost on it hah.

Also, I think I can do part C if I figure out part B, but any help would still be appreciated.

I can't show the picture but it is a solid disk with the properties listed, with a counterweight hanging down perpendicular to the radius of the disk. If you have any questions about how it looks I can answer them. Thanks in advance.
1. The problem statement, all variables and given/known data



2. Relevant equations



3. The attempt at a solution
 

tiny-tim

Science Advisor
Homework Helper
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Hi Guruu! :smile:

(why did you keep using w, instead of copying the ω in the question? :wink:)
(b) When the counterweight has a speed v, the pulley has an angular speed ω = v/R. Determine the total angular momentum of the system about O.

My question involves part B. My best try was to calculate the angular momentum of the block and apply that to the whole system but that obviously doesn't work. My problem is with factoring in the counterweight when finding the angular momentum of hte whole system.
No, you must do the block and the cylinder separately.

The cylinder,as you know, is Iω, but the block is just the fundamental formula moment of inertia = speed times perpendicular distance (in other words it's the moment of the velocity). :smile:
 
Thanks a lot man.
 

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