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    How can I do 5 ii ?Attachment 727854
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    (Original post by joyoustele)
    How can I do 5 ii ?
    Label the diagram a bit more - include dots where appropriate to represent the centre of mass for each part of the rod. Secondly, measure out their distances along their respectively parts. Furthermore, the wire is uniform, so it has the same density throughout which you can call \mu. Then the mass of each part of the rod is just m=\mu \ell where \ell is the length of the part.

    Also, label \angle BCD = \theta and so you know that \cos \theta = \frac{4}{5}. Use this to determine \sin \theta as you'll be using these.

    So then begin the question, finding the horizontal distance of the overall CoM from the point B. This is the same as finding the horizontal distance from the part AB. You have 3 different objects each with their own CoM so just determine the overall one from a formula you know.
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    (Original post by RDKGames)
    Label the diagram a bit more - include dots where appropriate to represent the centre of mass for each part of the rod. Secondly, measure out their distances along their respectively parts. Furthermore, the wire is uniform, so it has the same density throughout which you can call \mu. Then the mass of each part of the rod is just m=\mu \ell where \ell is the length of the part.

    Also, label \angle BCD = \theta and so you know that \cos \theta = \frac{4}{5}. Use this to determine \sin \theta as you'll be using these.

    So then begin the question, finding the horizontal distance of the overall CoM from the point B. This is the same as finding the horizontal distance from the part AB. You have 3 different objects each with their own CoM so just determine the overall one from a formula you know.
    Thanks very much
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    (Original post by RDKGames)
    Label the diagram a bit more - include dots where appropriate to represent the centre of mass for each part of the rod. Secondly, measure out their distances along their respectively parts. Furthermore, the wire is uniform, so it has the same density throughout which you can call \mu. Then the mass of each part of the rod is just m=\mu \ell where \ell is the length of the part.

    Also, label \angle BCD = \theta and so you know that \cos \theta = \frac{4}{5}. Use this to determine \sin \theta as you'll be using these.

    So then begin the question, finding the horizontal distance of the overall CoM from the point B. This is the same as finding the horizontal distance from the part AB. You have 3 different objects each with their own CoM so just determine the overall one from a formula you know.
    Can you tell me why sometimes the y is ignored?
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    (Original post by joyoustele)
    Can you tell me why sometimes the y is ignored?
    What's y?
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    (Original post by RDKGames)
    What's y?
    y bar. Nm I think I know
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    (Original post by joyoustele)
    y bar. Nm I think I know
    I assume that \bar{y} is the vertical height of the CoM. So of course you ignore this for part (i) since you're doing it horizontally here, and then you do \bar{y} in part (ii)
 
 
 
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