M5 June 2014 Q6

(a) Prove, using integration, that the moment of inertia of a uniform circular disc, of mass \(m\) and radius \(a\), about an axis through the centre of the disc and perpendicular to the plane of the disc is \(\dfrac{1}{2}ma^2\). (5)

[You may assume without proof that the moment of inertia of a uniform hoop of mass \(m\) and radius \(r\) about an axis through its centre and perpendicular to its plane is \(mr^2\).]

Figure 1: annulus with centre O, inner radius a and outer radius 2a — Figure 1

A uniform plane shape \(S\) of mass \(M\) is formed by removing a uniform circular disc with centre \(O\) and radius \(a\) from a uniform circular disc with centre \(O\) and radius \(2a\), as shown in Figure 1. The shape \(S\) is free to rotate about a fixed smooth axis \(L\), which passes through \(O\) and lies in the plane of the shape.

(b) Show that the moment of inertia of \(S\) about \(L\) is \(\dfrac{5}{4}Ma^2\). (4)

The shape \(S\) is at rest in a horizontal plane and is free to rotate about the axis \(L\). A particle of mass \(M\) falls vertically and strikes \(S\) at the point \(A\), where \(OA = \dfrac{3}{2}a\) and \(OA\) is perpendicular to \(L\). The particle adheres to \(S\) at \(A\). Immediately before the particle strikes \(S\) the speed of the particle is \(u\).

Mark scheme (a)

Scheme	Marks
\(\delta m = 2\pi x\delta x\dfrac{m}{\pi a^2} = \dfrac{2mx\delta x}{a^2}\)	M1 A1
\(\delta I = \dfrac{2mx^3\delta x}{a^2}\)	A1
\(I = \dfrac{2m}{a^2}\displaystyle\int_0^a x^3\,\mathrm{d}x\)	M1
\(= \tfrac{1}{2}ma^2\) PRINTED	A1
	(5)

Notes

First M1 for area element

First A1 for a correct \(\delta m\)

Second A1 for a correct \(\delta I\)

Second M1 for using mass per unit area and integrating with correct limits

Third A1 for the PRINTED ANSWER

Mark scheme (b)

Scheme	Marks
\(\tfrac{1}{2}\dfrac{4M}{3}(2a)^2 - \tfrac{1}{2}\dfrac{M}{3}a^2\)	M1 A1
\(= \tfrac{5}{2}Ma^2\)
\(2I = \tfrac{5}{2}Ma^2\) (perp axes rule)	M1
\(I = \tfrac{5}{4}Ma^2\) PRINTED	A1
	(4)

Notes

M1 for use of difference of MI (difference in the masses must be \(M\))

A1 for correct expression without mass per unit area

M1 for use of MI about diameter (in formula book)

A1 for the PRINTED ANSWER

N.B. The two M marks may be earned in either order

Mark scheme (c)

Scheme	Marks
\(\left(\tfrac{5}{4}Ma^2 + M\left(\tfrac{3a}{2}\right)^2\right)\omega = Mu\left(\tfrac{3a}{2}\right)\)	M1 A1 A1
\(\omega = \tfrac{3u}{7a}\)	A1
KE loss \(= \tfrac{1}{2}Mu^2 - \tfrac{1}{2}\left(\tfrac{5}{4}Ma^2 + M\left(\tfrac{3a}{2}\right)^2\right)\left(\tfrac{3u}{7a}\right)^2\)	M1 A2 ft
\(= \dfrac{5Mu^2}{28}\)	A1
	(8)
	(17 marks)

Notes

First M1 for conservation of angular momentum equation

First A1 for LHS on scheme

Second A1 for RHS on scheme

Third A1 for a correct \(\omega\) (or possibly \(v\))

Second M1 for a difference in KE (must have found an \(\omega\))

(omission of MI of particle is missing term so M0)

Fourth and fifth A2 ft on their \(\omega\)

Sixth A1 for a correct positive answer