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Pulley/Torque problem where pulley has mass

  • Thread starter tron_2.0
  • Start date
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hi, im new to the forum =] my names suraj and im an engineering/journalism student in my first year. this forum looked pretty informative so i figured id sign up. anyways id really appreciate it if someone could help me out on this problem.

1. Homework Statement

Consider two objects with connected by a light string that passes over a pulley having a moment of inertia of about its axis of rotation as shown in the figure below. The string does not slip on the pulley or stretch. The pulley turns without friction. The two objects are released from rest separated by a vertical distance .

(a) Use the principle of conservation of energy to find the translational speeds of the objects as they pass each other.
(b) Find the angular speed of the pulley at this time.


2. Homework Equations
well not given in the problem but i believe you have to apply:
-Free Body Diagrams (of both masses)
-Torque=Fd
-Kinematics

3. The Attempt at a Solution

okay well i tried to solve it in the variable form that the question is asking for, and i was unable to do so. however i tried to plug in some numbers (give each mass a numerical mass, give the pulley a random mass, etc) and i believe i solved the problem. im just confused as to how i should solve it in the form using variables?


this is what i got when i made up values for the pulley's mass, the pulley's radius, the mass of the weight on the left, and the mass of the weight on the right:

http://i21.photobucket.com/albums/b277/riceboy89/solution1.jpg [Broken]

however it is asking me the answer in terms of variables, and im having a hard time trying to find my acceleration in terms of variables so i can use the equation i ended up in the above solution (i figured that if i can solve for a in terms of variables, cant i just plug it into v=sqrt(2a*deltaY)?)

http://i21.photobucket.com/albums/b277/riceboy89/solution2.jpg [Broken]

thanks for the guidance
 
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Answers and Replies

Hi Suraj!

In http://i21.photobucket.com/albums/b277/riceboy89/solution2.jpg" [Broken] pic, as

[tex]m_1>m_2[/tex] the equations should be

[tex]m_1g-T_1=m_1a[/tex]

and

[tex]
T_2-m_2g=m_2a[/tex]

The [tex]\tau=I\alpha[/tex] equation looks good to me.
 
Last edited by a moderator:
335
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You will have a loss in PE balanced by a gain in KE and work done.

What is it that loses PE, and by how much ?

What are the gains in KE and by how much ?

What work is done and how much ?
 
29
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google_spider:

am i on the right track in finding the acceleration first, in terms of variables, and then plugging it into the equation (sqrt(2a*deltaY))?
 
am i on the right track in finding the acceleration first, in terms of variables, and then plugging it into the equation (sqrt(2a*deltaY))?
What is [tex]a_y=\sqrt{2a\delta y}[/tex] ?? I have never come across this formula. As Phisixguru pointed out, you have to use energy conservation. Both the blocks will be at a height of 4H when they pass each other.
 
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