M3 June 2007 Q4

EdexcelOld spec9 marksHorizontal Circular Motion

4. A light inextensible string of length \(l\) has one end attached to a fixed point \(A\). The other end is attached to a particle \(P\) of mass \(m\). The particle moves with constant speed \(v\) in a horizontal circle with the string taut. The centre of the circle is vertically below \(A\) and the radius of the circle is \(r\).

Show that

\[gr^2 = v^2\sqrt{(l^2 - r^2)}.\]

(9)

Diagram: string AP of length l at angle theta to the vertical through A, tension T along the string, weight mg at P, horizontal circle of radius r — Mark scheme

Scheme	Marks
\(\uparrow\) \(T\cos\theta = mg\)	M1 A1
\(\leftarrow\) \(T\sin\theta = \dfrac{mv^2}{r}\)	M1 A1
\(\tan\theta = \dfrac{r}{\sqrt{\left(l^2 - r^2\right)}}\) or equivalent	M1 A1
\(\tan\theta = \dfrac{v^2}{rg}\) Eliminating \(T\)	M1
\(\dfrac{r}{\sqrt{\left(l^2 - r^2\right)}} = \dfrac{v^2}{rg}\) Eliminating \(\theta\)	M1
\(gr^2 = v^2\sqrt{\left(l^2 - r^2\right)}\) * cso	A1
	(9)
	(9 marks)