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    I've attempted this question so many times and my answer is always slightly off. I'm unsure what I'm doing wrong. Please could someone have a go?

    I've attached a picture with the necessary information. The rod was held horizontally and released from rest. The question asks to find the linear velocity of A at the bottom of its path. According to the textbook, the correct answer is 1.46m/s. Thanks

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    (Original post by PhyM23)
    I've attempted this question so many times and my answer is always slightly off. I'm unsure what I'm doing wrong. Please could someone have a go?

    I've attached a picture with the necessary information. The rod was held horizontally and released from rest. The question asks to find the linear velocity of A at the bottom of its path. According to the textbook, the correct answer is 1.46m/s. Thanks
    Is it possible that you post your working?
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    (Original post by Eimmanuel)
    Is it possible that you post your working?
    Of course. Here you go
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    (Original post by PhyM23)
    Of course. Here you go
    Can clarify about the length of the rod:
    in the first post, the separation between the mass A and B is l but in the 3rd posting, the separation becomes 2l.
    So which is the correct data?
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    (Original post by Eimmanuel)
    Can clarify about the length of the rod:
    in the first post, the separation between the mass A and B is l but in the 3rd posting, the separation becomes 2l.
    So which is the correct data?
    Yes my apologies. The length of the rod is 0.8m. I used different notation on different occasions which is where the confusion lies. So in the second case, 2l=0.8m so l=0.4m
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    (Original post by PhyM23)
    Of course. Here you go
    It seems that you are mixing two different speeds together: the speed of the rod and mass.

    I suggest you work on the angular speed omega and use it to compute the linear speed of mass A.

    I can get 1.46 m/s.
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    (Original post by Eimmanuel)
    It seems that you are mixing two different speeds together: the speed of the rod and mass.

    I suggest you work on the angular speed omega and use it to compute the linear speed of mass A.

    I can get 1.46 m/s.
    Thank you for having a go. I've tried that I believe but I think I'm confused which energies I should use. Please may you show your working?
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    (Original post by PhyM23)
    Thank you for having a go. I've tried that I believe but I think I'm confused which energies I should use. Please may you show your working?
    Change in PE = Change in KE

    Change in PE =  (m_A - m_B)g(\frac{l}{2})

    Change in KE =  \frac{1}{2}m_A (\frac{l}{2})^2 \omega^2 + \frac{1}{2}m_B (\frac{l}{2})^2 \omega^2 + \frac{1}{2} I_{rod} \omega^2

     I_{rod} = \frac{1}{12} M_{rod} l^2

    I am using the first post info.
     l = 0.8 m

    You should be able to use the two equations to find  \omega and find the linear speed.
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    (Original post by Eimmanuel)
    Change in PE = Change in KE

    Change in PE =  (m_A - m_B)g(\frac{l}{2})

    Change in KE =  \frac{1}{2}m_A (\frac{l}{2})^2 \omega^2 + \frac{1}{2}m_B (\frac{l}{2})^2 \omega^2 + \frac{1}{2} I_{rod} \omega^2

     I_{rod} = \frac{1}{12} M_{rod} l^2

    I am using the first post info.
     l = 0.8 m

    You should be able to use the two equations to find  \omega and find the linear speed.
    May I ask why you use the moment of inertia of the rod and not the whole system?
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    (Original post by PhyM23)
    May I ask why you use the moment of inertia of the rod and not the whole system?
    How do you find the moment of inertia of the whole system?
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    (Original post by Eimmanuel)
    How do you find the moment of inertia of the whole system?
    I did I_{rod} + m_{A}l^{2} + m_{B}l^{2}

    Where l is the distance from each mass to the centre.
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    (Original post by PhyM23)
    I did I_{rod} + m_{A}l^{2} + m_{B}l^{2}

    Where l is the distance from each mass to the centre.
    Then you are double counting the KE of A and B based on your working.
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    (Original post by Eimmanuel)
    Then you are double counting the KE of A and B based on your working.
    Ah yes so I am. Thank you very much for your help; I really appreciate it.
 
 
 
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