M3 June 2015 Q3

EdexcelOld spec10 marksHorizontal Circular Motion

Figure 1: ball P of mass m attached by strings 2l to A above and B below, AP at 30 degrees to AB, rotating with angular speed omega — Figure 1

A small ball \(P\) of mass \(m\) is attached to the midpoint of a light inextensible string of length \(4l\). The ends of the string are attached to fixed points \(A\) and \(B\), where \(A\) is vertically above \(B\). Both strings are taut and \(AP\) makes an angle of 30\(^\circ\) with \(AB\), as shown in Figure 1. The ball is moving in a horizontal circle with constant angular speed \(\omega\).

(a) Find, in terms of \(m\), \(g\), \(l\) and \(\omega\),

(i) the tension in \(AP\),

(ii) the tension in \(BP\). (8)

(b) Show that \(\omega^2 \geqslant \dfrac{g\sqrt{3}}{3l}\). (2)

Mark scheme (a)

Scheme	Marks
R\((\uparrow)\) \(T_A\cos 30 = mg + T_B\cos 30\)	M1A1
NL2 \(T_A\cos 60 + T_B\cos 60 = mr\omega^2\)	M1A1
\(= m \times 2l\cos 60\,\omega^2\) or \(ml\omega^2\)	A1
\(T_A + T_B = 2ml\omega^2\)
\(T_A - T_B = \dfrac{2mg}{\sqrt{3}}\)
(i) \(T_A = \dfrac{m}{3}\left(3l\omega^2 + g\sqrt{3}\right)\) oe	DM1A1
(ii) \(T_B = \dfrac{m}{3}\left(3l\omega^2 - g\sqrt{3}\right)\) oe	A1
	(8)

Notes

M1 Resolving vertically

A1 Correct equation

M1 NL2 along radius, acceleration in either form

A1 LHS correct

A1 Correct radius substituted and accel in \(r\omega^2\). Can be awarded later by implication if work implies correct radius used.

DM1 Solving the two equations to obtain an expression for either tension. Depenent on both previous M marks

A1 Tension in \(AP\) correct – simplified to two terms

A1 Tension in \(BP\) correct – simplified to two terms

Mark scheme (b)

Scheme	Marks
\(T_B \geqslant 0 \ \Rightarrow\ 3l\omega^2 \geqslant g\sqrt{3}\)	M1
\(\omega^2 \geqslant \dfrac{g\sqrt{3}}{3l}\) *	A1cso
	(2)
	(10 marks)

Notes

M1 Using their tension in \(BP \geqslant 0\) must be \(\geqslant\) for this mark

A1cso Obtaining the GIVEN answer. Only error allowed is the expression for the tension in \(AP\)