M3 June 2008 Q3

EdexcelOld spec13 marksHorizontal Circular Motion

Figure 2: string from A to B at angle theta to the vertical AO, B on the horizontal table, AO = h — Figure 2

Figure 2 shows a particle \(B\), of mass \(m\), attached to one end of a light elastic string. The other end of the string is attached to a fixed point \(A\), at a distance \(h\) vertically above a smooth horizontal table. The particle moves on the table in a horizontal circle with centre \(O\), where \(O\) is vertically below \(A\). The string makes a constant angle \(\theta\) with the downward vertical and \(B\) moves with constant angular speed \(\omega\) about \(OA\).

(a) Show that \(\omega^2 \leqslant \dfrac{g}{h}\). (8)

The elastic string has natural length \(h\) and modulus of elasticity \(2mg\).

Given that \(\tan\theta = \dfrac{3}{4}\),

(b) find \(\omega\) in terms of \(g\) and \(h\). (5)

Force diagram: N up, T along the string at angle theta, mg down, acceleration r omega squared horizontally — Mark scheme (a)

Right-angled triangle with hypotenuse l, vertical side h, horizontal side r, angle theta at the top — Mark scheme (a)

Scheme	Marks
\(\uparrow\) \(T\cos\theta + N = Mg\) (1)	M1A1
\(\rightarrow\) \(T\sin\theta = mr\omega^2\) (2)	M1A1
\(\sin\theta = \dfrac{r}{l}\) from (2) \(T = ml\omega^2\)
sub into (1) \(ml\cos\theta\,\omega^2 + N = mg\)	M1
\(N = mg - mh\omega^2\)	A1
Since in contact with table \(N \geqslant 0\) \(\therefore \omega^2 \leqslant \dfrac{g}{h}\) *	M1A1 cso
	(8)

Mark scheme (b)

Scheme	Marks
\(r : h : l = 3 : 4 : 5 \quad \therefore\) extension \(= \dfrac{h}{4}\)	B1
\(T = \dfrac{2mg}{h} \times \dfrac{h}{4} = \dfrac{mg}{2}\)	M1A1
\(T = ml\omega^2 = \dfrac{5mh}{4}\omega^2 \quad \omega = \sqrt{\dfrac{2g}{5h}}\)	M1A1
	(5)
	(13 marks)