M5 June 2015 Q7

(a) Find, using integration, the moment of inertia of a uniform solid hemisphere, of mass \(m\) and radius \(a\), about a diameter of its plane face.

[You may assume, without proof, that the moment of inertia of a uniform circular disc, of mass \(m\) and radius \(r\), about a diameter is \(\dfrac{1}{4}mr^2\).] (10)

(b) Hence find the moment of inertia of a uniform solid sphere, of mass \(M\) and radius \(a\), about a diameter. (2)

Mark scheme (a)

Scheme	Marks
\(\delta V = \pi y^2\delta x\)	M1
\(\delta m = \pi y^2\delta x\dfrac{3m}{2\pi a^3}\)	M1
\(\delta I = \tfrac{1}{4}\delta m y^2 + \delta m x^2\)	M1 M1 A1
\(= \tfrac{1}{4}\delta m(y^2 + 4x^2)\)
\(= \tfrac{1}{4}\pi(a^2 - x^2)\delta x\dfrac{3m}{2\pi a^3}(a^2 - x^2 + 4x^2)\)	M1
\(= \tfrac{1}{4}\pi(a^2 - x^2)(a^2 + 3x^2)\delta x\dfrac{3m}{2\pi a^3}\)
\(= \dfrac{3m}{8a^3}(a^4 + 2a^2x^2 - 3x^4)\delta x\)	A1
\(I = \dfrac{3m}{8a^3}\displaystyle\int_0^a (a^4 + 2a^2x^2 - 3x^4)\,\mathrm{d}x\)	M1
\(= \dfrac{3m}{8a^3}\left[a^4x + \dfrac{2a^2x^3}{3} - \dfrac{3x^5}{5}\right]_0^a\)	A1
\(= \dfrac{2ma^2}{5}\)	A1
	(10)

Notes

First M1 for vol. element

Second M1 for their \(\delta V \times\) correct density

Third M1 for \(\tfrac{1}{4}\delta m y^2\)

Fourth M1 for use of parallel axes

First A1 for a correct expression in terms of \(x\), \(y\) and \(\delta m\)

Fifth M1 for sub for \(\delta m\) and \(y\)

Second A1 for a correct \(\delta I\) in terms of \(x\) only

Sixth M1 for integrating with correct limits

Second A1 for correct integral

Third A1 for the answer

Mark scheme (b)

Scheme	Marks
\(I = 2 \times 2 \times \left(\dfrac{1}{2}M\right)\dfrac{a^2}{5}\)	M1
\(= \dfrac{2Ma^2}{5}\)	A1
	(2)
	(12 marks)

Notes

M1 for use of additive rule with adjusted mass

A1 for correct answer

N.B. No marks for non-hence method