M5 June 2005 Q4

Figure 1: rod PQ of length 3a hanging from P on horizontal axis L, with disc of radius a attached at Q, diameter QR in line with the rod — Figure 1

A thin uniform rod \(PQ\) has mass \(m\) and length \(3a\). A thin uniform circular disc, of mass \(m\) and radius \(a\), is attached to the rod at \(Q\) in such a way that the rod and the diameter \(QR\) of the disc are in a straight line with \(PR = 5a\). The rod together with the disc form a composite body, as shown in Figure 1. The body is free to rotate about a fixed smooth horizontal axis \(L\) through \(P\), perpendicular to \(PQ\) and in the plane of the disc.

(a) Show that the moment of inertia of the body about \(L\) is \(\dfrac{77ma^2}{4}\). (7)

When \(PR\) is vertical, the body has angular speed \(\omega\) and the centre of the disc strikes a stationary particle of mass \(\tfrac{1}{2}m\). Given that the particle adheres to the centre of the disc,

(b) find, in terms of \(\omega\), the angular speed of the body immediately after the impact. (4)

Sketch: rod of length 3a from P on axis L to Q, disc of radius a below with diameter QR — Mark scheme (a)

Scheme	Marks
DISC: \(I_{\text{diam}} = \dfrac{1}{2}\left(\dfrac{1}{2}ma^2\right) = \dfrac{1}{4}ma^2\)	M1 A1
\(I_L = \dfrac{1}{4}ma^2 + m(4a)^2\)	M1 A1
\(= \dfrac{65}{4}ma^2\)
ROD: \(I_L = \dfrac{4}{3}m\left(\dfrac{3a}{2}\right)^2 = 3ma^2\)	B1
\(I_{\text{TOTAL}} = \dfrac{65}{4}ma^2 + 3ma^2 = \dfrac{77}{4}ma^2\) *	M1 A1
	(7)

Scheme	Marks
CAM: \(\dfrac{77}{4}ma^2\omega = \left[\dfrac{77}{4}ma^2 + \dfrac{1}{2}m(4a)^2\right]\omega'\)	M1 A1 A1
\(\Rightarrow\ \omega' = \dfrac{77}{109}\omega\)	A1
	(4)
	(11 marks)