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Angular speed

  1. Oct 25, 2009 #1
    1. The problem statement, all variables and given/known data


    The pulley in system drawn below has radius 8.0 cm and mass 0.6 kg. The surface on which the 4.0 kg mass is resting is frictionless. Suppose the 2.0 kg block is released from rest.

    http://bcs.whfreeman.com/WebPub/Physics/tiplerphysics6e_bridge/question_bank_images/9-72.png?9954


    1. What is the speed of the block in m/s after it falls a distance of 2.5 m. Round your answer to one decimal place. Derive a complete formula for the answer before substituting values to avoid round-off errors. Use g = 9.81 m/s^2.



    2. What is the angular speed of the pulley in rad/s at the instant the block reaches a distance of 2.5 m? Round your answer to one decimal place. Derive a complete formula for the answer before substituting values to avoid round-off errors. Use g = 9.81 m/s^2. Round your answer to two significant figures.



    2. Relevant equations



    3. The attempt at a solution

    What I did is:

    T1=m1a (T1 is tension connecting the 4 kg block, m1 is the mass of 4 kg block)

    (T2-T1) = I [tex]\alpha[/tex] ( T2 is tension connecting the 2 kg block, I is the moment of inertia of pulley)

    m2g - T2 = m2a (m2 is mass of 2 kg block)

    These are the three equations I was able to find but how do I get the answers to these question?
     
  2. jcsd
  3. Oct 25, 2009 #2

    Delphi51

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    You have 4 unknowns, so you'll need another equation.
    Can you relate "a" and "alpha"? You know the radius . . .
     
  4. Oct 25, 2009 #3
    at=r alpha?
     
  5. Oct 25, 2009 #4

    Delphi51

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    Okay, you should have 4 equations and 4 unknowns now, so just a matter of substitution to solve for the one you want.
     
  6. Oct 25, 2009 #5
    First question asks for speed of the block after 2.5 m

    I used the conservation of energy equation. I got mghi = 0.5 mv^2f

    So I got vf = 7, is it right?

    I still can't figure out the second question..
     
  7. Oct 25, 2009 #6

    Delphi51

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    mghi = 0.5 mv^2f doesn't include the rotational energy of the pulley.
    You'll have to help me - I forget how to find the moment of inertia I for the pulley. It is just a cylinder, so you probably have a formula for it.

    For #2, do you have a formula relating the velocity to the angular velocity?
     
  8. Oct 25, 2009 #7
  9. Oct 26, 2009 #8

    Delphi51

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    Okay, so you can find the moment of inertia of the pulley and add its rotational energy term to your equation.
     
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