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Rolling hemispheres and moments of inertia

  1. Apr 7, 2009 #1
    1. The problem statement, all variables and given/known data

    Hi there - I'm working through a bit of mechanics revision and have been looking at some of my uni past exam papers from a good few years back when the syllabus was different, so I'm not familiar with a lot of the material covered - it's not strictly relevant to my current course but I've found myself stuck on this question about hemispheres and it's bugging me, relevant or not. Could someone please give me a little help with how to progress? Thanks a lot.

    i) Use the parallel axis theorem to calculate the moment of inertia of a uniform hemisphere of mass 'M' and radius 'a' about an axis through its centre of mass and parallel to the base (the centre of mass is located at a distance of 3a/8 from the flat face of the hemisphere).

    Well the moment of inertia of the hemisphere about the axis which would be through the centre of mass of a full sphere is [itex]\frac{2ma^2}{5}[/itex], as it is for a sphere, so by the parallel axes theorem, [itex]I_{c.o.m}+M(\frac{3a}{8})^2=\frac{2Ma^2}{5}[/itex] so [itex]I_{c.o.m}=Ma^2(\frac{2}{5}-\frac{9}{64})=\frac{83}{320}Ma^2[/itex]. Now for a start, 83/320 seems like about the ugliest moment of inertia coefficient I've ever seen, so I'm pretty sure I've gone wrong somewhere here - but where?

    ii) The hemisphere initially rests on a rough horizontal plane with its base vertical. It is then released from rest and rolls on the plane without slipping. Let [itex]\theta[/itex] be the angle that the base makes with the horizontal at time t - express the instantaneous speed of the centre of mass in terms of b and the rate of change of [itex]\theta[/itex], where b is the instantaneous distance from the centre of mass to the point of contact with the plane. Hence write down expressions for the potential and kinetic energy of the hemisphere and deduce that [itex](\frac{d \theta}{dt})^2=\frac{15gcos\theta}{(28-15cos\theta)a}[/itex].

    Now I really have no idea what on earth to do on this one, and I can see that part (i) is probably relevant but it's not rotating around the centre of mass ('C') so I can only assume they want us to calculate another moment of inertia which is just plain annoying. The picture I have has that the midpoint of the 'base edge' (where the centre of mass of a full sphere would be - call this point 'P', say) is always directly above the point of contact of the sphere and the surface - call this 'B' say - I'm pretty sure this is correct because at the very least it's true at the start and the end of the motion. In that case, by the cosine rule for the triangle CPB, the length of side CB squared is [itex]b^2=a^2 + \frac{9a^2}{64}-2(a)(\frac{3a}{8})cos(\theta)=a^2(\frac{73-48cos\theta}{64})[/itex] which again seems wrong. Even if that is the correct expression for [itex]b^2[/itex], I'm not sure how to express the speed of the centre of mass in terms of [itex]b[/itex] and [itex]\dot \theta[/itex], and will the kinetic energy expression include an [itex]\frac{I{\dot \theta}^2}{2}[/itex] or an [itex]\frac{mv^2}{2}[/itex] or both? Apologies for my cluelessness, as I said this is a fair bit off the syllabus I'm meant to know! Thanks a lot,

    1. The problem statement, all variables and given/known data

    2. Relevant equations

    3. The attempt at a solution
  2. jcsd
  3. Apr 7, 2009 #2
    For Part i you appoached the problem correctly. I'll look at ii. Are you familiar with Lagangian Mechanics?
  4. Apr 7, 2009 #3
    I don't believe so but if you point me in the right direction I'll be happy to read up! So the moment of inertia -is- that ugly thing?
  5. Apr 7, 2009 #4
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