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# M5 June 2010 Q3a Watch

1. 3. A uniform lamina ABC of mass m is in the shape of an isosceles triangle with
AB = AC = 5a and BC = 8a. (a) Show, using integration, that the moment of inertia of the lamina about an axisthrough A, parallel to BC, is 92 ma2 .
The answer should read (9/2)ma^2. In the mark scheme for this question it makes no reference to the use of the standard result for a uniform rod of length 2l, and neither does the question, and yet surely this is essential since you need the summation of all the strip's moment of inertia to get the total moment of inertia.

2. (Original post by mathsmusician)
3. A uniform lamina ABC of mass m is in the shape of an isosceles triangle with
AB = AC = 5a and BC = 8a. (a) Show, using integration, that the moment of inertia of the lamina about an axisthrough A, parallel to BC, is 92 ma2 .
The answer should read (9/2)ma^2. In the mark scheme for this question it makes no reference to the use of the standard result for a uniform rod of length 2l, and neither does the question, and yet surely this is essential since you need the summation of all the strip's moment of inertia to get the total moment of inertia.

The "standard result" for the M of I of a rod is about an axis perpendicular to the rod. That is not needed in this question (or at least, the way the mark scheme and I have done it). The lamina is split into strips which are parallel to BC. All of the typical strip is a distance x from A, so its M of I about an axis through A , parallel to BC is mass of strip times x^2.
3. Okay thanks that makes sense. What I am wondering now is why can't this method always be applied....why in some cases do you have to use the standard result for the M.I. of a rod to work out the M.I. of the lamina. For example in June 2013 Q5 there's an almost identical problem, the only difference is that the axis of rotation is perpendicular to the base of the triangle, not parallel to it as in this question. For this type of question, couldn't you simply use the same method without using a standard result, and use the perpendicular rule to get the final answer?
4. (Original post by mathsmusician)
Okay thanks that makes sense. What I am wondering now is why can't this method always be applied....why in some cases do you have to use the standard result for the M.I. of a rod to work out the M.I. of the lamina. For example in June 2013 Q5 there's an almost identical problem, the only difference is that the axis of rotation is perpendicular to the base of the triangle, not parallel to it as in this question. For this type of question, couldn't you simply use the same method without using a standard result, and use the perpendicular rule to get the final answer?
When you use perpendicular axes for a circle, for instance, you use the symmetry of the circle. You haven't got that symmetry for the triangle, as I see it.
5. Yep that answers my question. Thanks

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