CIE Further Maths Moment of Inertia

Maths and statistics discussion, revision, exam and homework help.

This thread is sponsored by:
Announcements Posted on
Important: please read these guidelines before posting about exams on The Student Room 28-04-2013
Sign in to Reply
  1. johnconnor92's Avatar
    • Respected Member
    • Location: Viridian City
    • Posts: 229
    CIE Further Maths Moment of Inertia
    Click image for larger version. 

Name:	questions.png 
Views:	32 
Size:	61.8 KB 
ID:	145700
    Hey guys, I really need your help here. I tried calculating the MI of the object by first summing the sphere's and the particles about the axis C, and then moving both of them to the axis O. by parallel axis theorem it would look like
    [M(4a^2) + 2(3M)(2a)^2/5] + 4M (a^2)

    which gives me (64Ma^2)/5. What am I missing? Isn't this method valid?

    Click image for larger version. 

Name:	answer.png 
Views:	24 
Size:	70.2 KB 
ID:	145701
    Also, I notice that the marking scheme for part (i) is kind of weird. Why did the "3" become "6" all of a sudden after the approximation?

    Please help. Thank you.
  2. ghostwalker's Avatar
    • Outcast of Imrryr
    • Location: CA13
    Re: CIE Further Maths Moment of Inertia
    (Original post by johnconnor92)

    which gives me (64Ma^2)/5. What am I missing? Isn't this method valid?
    Axis C is not the centre of mass of sphere and particle combination, so you can't apply the parallel axes theorem on the combined mass.

    Also, I notice that the marking scheme for part (i) is kind of weird. Why did the "3" become "6" all of a sudden after the approximation?
    The first equation in answer to part (i) is spread over two lines, with two terms in all, each containing -3...., hence on addition -6.
    Last edited by ghostwalker; 04-05-2012 at 08:43.
  3. johnconnor92's Avatar
    • Respected Member
    • Location: Viridian City
    • Posts: 229
    Re: CIE Further Maths Moment of Inertia
    (Original post by ghostwalker)
    Axis C is not the centre of mass of sphere and particle combination, so you can't apply the parallel axes theorem on the combined mass.
    Why not? In what cases, then, am I able to apply parallel axes theorem on composite objects?
  4. ghostwalker's Avatar
    • Outcast of Imrryr
    • Location: CA13
    Re: CIE Further Maths Moment of Inertia
    (Original post by johnconnor92)
    Why not? In what cases, then, am I able to apply parallel axes theorem on composite objects?
    When you know the centre of mass. Check out the parallel axes theorem, it doesn't work on any old parallel axes.
  5. johnconnor92's Avatar
    • Respected Member
    • Location: Viridian City
    • Posts: 229
    Re: CIE Further Maths Moment of Inertia
    (Original post by ghostwalker)
    When you know the centre of mass. Check out the parallel axes theorem, it doesn't work on any old parallel axes.
    So I can only apply the parallel axes theorem if I first find out the centre of mass of the two objects (which is somewhere below O)?
  6. ghostwalker's Avatar
    • Outcast of Imrryr
    • Location: CA13
    Re: CIE Further Maths Moment of Inertia
    (Original post by johnconnor92)
    So I can only apply the parallel axes theorem if I first find out the centre of mass of the two objects (which is somewhere below O)?
    If you're working on the combined masses together, then yes.


    Which is why the mark scheme works on the individual components.
Sign in to Reply
Share this discussion:  
Article updates
Moderators

We have a brilliant team of more than 60 volunteers looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.

Reputation gems:
The Reputation gems seen here indicate how well reputed the user is, red gem indicate negative reputation and green indicates a good rep.
Post rating score:
These scores show if a post has been positively or negatively rated by our members.