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    I've decided to self teach FP3 (MEI course) this year. I've just got the textbook and I'm working through some vector product questions. Sadly same worded questions that require proof don't have an accompanying solution at the end of the book (only final answers are given). Can anyone help I'm not sure where to start with these 3 questions (Q10-12), thanks!

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    (Original post by jonathanyyt)
    I've decided to self teach FP3 (MEI course) this year. I've just got the textbook and I'm working through some vector product questions. Sadly same worded questions that require proof don't have an accompanying solution at the end of the book (only final answers are given). Can anyone help I'm not sure where to start with these 3 questions (Q10-12), thanks!

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Views: 134
Size:  144.8 KBName:  ImageUploadedByStudent Room1421872105.742939.jpg
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    What definition if the cross product do they use in FP3?

    For Q10 note that a + b = -c (all vectors) so you should be able to do things like a x (a + b) = a x (-c) and follow the logic through from there.
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    (Original post by davros)
    What definition if the cross product do they use in FP3?

    For Q10 note that a + b = -c (all vectors) so you should be able to do things like a x (a + b) = a x (-c) and follow the logic through from there.
    They define it as \mathbf{a} X \mathbf{b} = |\mathbf{a}||\mathbf{b}|\sin \theta, and also \mathbf{a} X \mathbf{b} =\begin{pmatrix} a_1 \\a_2 \\a_3 \end{pmatrix} x \begin{pmatrix} b_1 \\b_2 \\b_3 \end{pmatrix} = \begin{pmatrix} a_2 b_3 - a_3b_2 \\a_3b_1 - a_1b_3 \\a_1b_2 - a_2b_1 \end{pmatrix}

    I know what you mean for Q10 now, How can I start Q11 and Q12?
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    (Original post by jonathanyyt)
    They define it as \mathbf{a} X \mathbf{b} = |\mathbf{a}||\mathbf{b}|\sin \theta, and also \mathbf{a} X \mathbf{b} =\begin{pmatrix} a_1 \\a_2 \\a_3 \end{pmatrix} x \begin{pmatrix} b_1 \\b_2 \\b_3 \end{pmatrix} = \begin{pmatrix} a_2 b_3 - a_3b_2 \\a_3b_1 - a_1b_3 \\a_1b_2 - a_2b_1 \end{pmatrix}

    I know what you mean for Q10 now, How can I start Q11 and Q12?
    Q11 is an extension of the idea used in Q10.

    Imagine marking another point S on the line of action of F so that vector OS = s. You want to prove that s x F = r x F i.e. it doesn't make a difference where you put point S as long as it's on the line of action of F.


    I haven't looked at Q12 in detail but I think you can argue it on physical princples given the definition you've just quoted to me for the cross product - in other words by thinking about what distance around a circle has been covered in a given time t using the angle made and the distance to the centre of the circle from a point on the axis.
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    (Original post by jonathanyyt)
    I've decided to self teach FP3 (MEI course) this year. I've just got the textbook and I'm working through some vector product questions. Sadly same worded questions that require proof don't have an accompanying solution at the end of the book (only final answers are given). Can anyone help I'm not sure where to start with these 3 questions (Q10-12), thanks!
    Q10: You can approach this geometrically - what is the area of the triangle in terms of the lengths of vectors a and b, b and c, c and a? (It relates easily to the magnitude of the cross product but you need to make sure that you take the cross products in the right order, else you'll end up with a sign error)

    Q12: Similarly, as davros suggested, you can find the magnitude of v easily in the form of the magnitude of a cross product. (You have a nice right triangle giving you the necessary \sin \theta). You then have to check that the direction generated by the cross product \bold{\omega} \times \bold{p} points in the direction of \bold v
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    (Original post by atsruser)
    Q10: You can approach this geometrically - what is the area of the triangle in terms of the lengths of vectors a and b, b and c, c and a? (It relates easily to the magnitude of the cross product but you need to make sure that you take the cross products in the right order, else you'll end up with a sign error)

    Q12: Similarly, as davros suggested, you can find the magnitude of v easily in the form of the magnitude of a cross product. (You have a nice right triangle giving you the necessary \sin \theta). You then have to check that the direction generated by the cross product \bold{\omega} \times \bold{p} points in the direction of \bold v
    I've managed Q10 and 11 now. I'm still stuck on Q12, perhaps I'm not visualising it well. Also the angular speed is rotating about an axis, but why is the velocity parallel to the axis? Thanks for taking your time to answer btw.


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    (Original post by jonathanyyt)
    Also the angular speed is rotating about an axis, but why is the velocity parallel to the axis?
    Your language is a bit sloppy here: "the angular speed is rotating about an axis" doesn't make much sense, and I'm not sure exactly what you mean.

    "but why is the velocity parallel to the axis?": by this I guess you mean "why does the angular velocity vector point along the axis of rotation?".

    If so, it's for two reasons:

    1. the axis of rotation is at right angles to the plane of rotation and hence uniquely defines the plane in which the rotation is taking place, and it's also constant (it's hard to describe the rotation in some other way with a vector; for example, the velocity vector of any point in the plane is always changing direction so don't make good candidates for describing a fixed quantity).

    2. when you point the vector in that direction, it makes the relationship \bold{v}=\bold{\omega} \times \bold{p} work out correctly as long as the velocity and angular velocity are related by a right hand grip rule (curl the fingers of your right hand the way the plane is turning, and your thumb points in the direction of the +ve angular velocity vector).
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    (Original post by atsruser)
    Your language is a bit sloppy here: "the angular speed is rotating about an axis" doesn't make much sense, and I'm not sure exactly what you mean.

    "but why is the velocity parallel to the axis?": by this I guess you mean "why does the angular velocity vector point along the axis of rotation?".

    If so, it's for two reasons:

    1. the axis of rotation is at right angles to the plane of rotation and hence uniquely defines the plane in which the rotation is taking place, and it's also constant (it's hard to describe the rotation in some other way with a vector; for example, the velocity vector of any point in the plane is always changing direction so don't make good candidates for describing a fixed quantity).

    2. when you point the vector in that direction, it makes the relationship \bold{v}=\bold{\omega} \times \bold{p} work out correctly as long as the velocity and angular velocity are related by a right hand grip rule (curl the fingers of your right hand the way the plane is turning, and your thumb points in the direction of the +ve angular velocity vector).
    Ok I understand why the vector is along the axis of rotation now. Apologies for my poor wording. I guess that point may be covered in mechanics modules that I don't do in school (only up to M2).

    How do I go about making that triangle you mentioned? I'm sorry I'm really poor at interpreting/manipulating diagrams


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    (Original post by jonathanyyt)
    How do I go about making that triangle you mentioned? I'm sorry I'm really poor at interpreting/manipulating diagrams
    They've already drawn the triangle in the 3D picture. It goes from O to the unnamed centre of the disc of rotation, to P.

    Call Q the centre of the disc of rotation and call \theta the angle QOP.

    What is the length PQ, in terms of \bold{p} and \theta?
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    (Original post by atsruser)
    They've already drawn the triangle in the 3D picture. It goes from O to the unnamed centre of the disc of rotation, to P.

    Call Q the centre of the disc of rotation and call \theta the angle QOP.

    What is the length PQ, in terms of \bold{p} and \theta?
    Am I right to say that PQ = |p|sin\theta then use v=rw to get the answer?


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