M3 June 2014 (R) Q2

EdexcelOld spec13 marksHorizontal Circular Motion

Figure 1: particle on the outer surface of a cone with vertex upwards, string of length l to a point above the vertex at 60 degrees to the horizontal, semi-vertical angle 60 degrees — Figure 1

A cone of semi-vertical angle 60\(^\circ\) is fixed with its axis vertical and vertex upwards. A particle of mass \(m\) is attached to one end of a light inextensible string of length \(l\). The other end of the string is attached to a fixed point vertically above the vertex of the cone. The particle moves in a horizontal circle on the smooth outer surface of the cone with constant angular speed \(\omega\), with the string making a constant angle 60\(^\circ\) with the horizontal, as shown in Figure 1.

(a) Find the tension in the string, in terms of \(m\), \(l\), \(\omega\) and \(g\). (7)

The particle remains on the surface of the cone.

(b) Show that the time for the particle to make one complete revolution is greater than \[2\pi\sqrt{\frac{l\sqrt{3}}{2g}}\] (6)

Mark scheme (a)

Scheme	Marks
\(T\sin 60^\circ + R\sin 60^\circ = mg\)	M1 A1
\(T\cos 60^\circ - R\cos 60^\circ = ml\cos 60^\circ\,\omega^2\)	M1 A1 A1
\(T = \tfrac{1}{2}m\left(l\omega^2 + \dfrac{2}{\sqrt{3}}g\right)\)	DM1 A1
	(7)

Notes

M1 vertical equation

A1 correct vertical equation

M1 horizontal equation, acceleration in either form

A1 correct lhs

A1 correct rhs

DM1 solve for \(T\)

A1 correct \(T\)

Mark scheme (b)

Scheme	Marks
\(R = \tfrac{1}{2}m\left(\dfrac{2}{\sqrt{3}}g - l\omega^2\right)\)	M1 A1
\(\tfrac{1}{2}m\left(\dfrac{2}{\sqrt{3}}g - l\omega^2\right) > 0\)	DM1
\(\omega < \sqrt{\dfrac{2g}{l\sqrt{3}}}\)	A1
\(t > 2\pi\sqrt{\dfrac{l\sqrt{3}}{2g}}\) **	DM1 A1
	(6)
	(13 marks)

Notes

M1 obtain an expression for \(R\)

A1 correct expression

DM1 setting \(R > 0\)

A1 correct inequality for w

DM1 obtaining an inequality for \(t\)

A1 correct inequality