M3 June 2005 Q5

EdexcelOld spec13 marksVertical Circular Motion

5. A smooth solid sphere, with centre \(O\) and radius \(a\), is fixed to the upper surface of a horizontal table. A particle \(P\) is placed on the surface of the sphere at a point \(A\), where \(OA\) makes an angle \(\alpha\) with the upward vertical, and \(0 < \alpha < \dfrac{\pi}{2}\). The particle is released from rest. When \(OP\) makes an angle \(\theta\) with the upward vertical, and \(P\) is still on the surface of the sphere, the speed of \(P\) is \(v\).

(a) Show that \(v^2 = 2ga(\cos\alpha - \cos\theta)\). (4)

Given that \(\cos\alpha = \tfrac{3}{4}\), find

(b) the value of \(\theta\) when \(P\) loses contact with the sphere, (5)

Diagram: sphere on the table, P on the surface at angle theta from the upward vertical, A at angle alpha, reaction R and weight mg at P, path of P leaving the sphere to B on the table — Mark scheme (a)

Scheme	Marks
\(\dfrac{1}{2}mv^2 = mg\left(\underline{a\cos\alpha - a\cos\theta}\right)\)	M1 A1 A1
\(v^2 = 2ga(\cos\alpha - \cos\theta)\) * cso	A1
	(4)

Mark scheme (b)

Scheme	Marks
\(mg\cos\theta\ (-R) = \dfrac{mv^2}{a} \quad (R = 0)\)	M1 A1=A1
\(g\cos\theta = 2g\left(\dfrac{3}{4} - \cos\theta\right)\)	M1
\(\cos\theta = \tfrac{1}{2} \Rightarrow \theta = \tfrac{\pi}{3}\) (accept \(60^\circ\))	A1
	(5)

Notes

The scheme brackets the two M1 marks: the second depends on the first.

Mark scheme (c)

Scheme	Marks
From \(A\) to \(B\) \(\dfrac{1}{2}mw^2 = mg\left(\underline{a + a\cos\alpha}\right)\)	M1 A1 A1
\(w^2 = 2ga\left(1 + \dfrac{3}{4}\right) \Rightarrow w = \left(\dfrac{7ga}{2}\right)^{\frac{1}{2}}\)	A1
	(4)
	(13 marks)

Alternative: from \(P\) to \(C\)

\(v_P^2 = 2ga\left(\dfrac{3}{4} - \dfrac{1}{2}\right) = \dfrac{ga}{2}\)
\(\dfrac{1}{2}mw^2 - \dfrac{1}{2}m\left(\dfrac{ga}{2}\right) = mg\left(\underline{a + a\cos\theta}\right)\)	M1 A1 A1
\(w^2 - \dfrac{ga}{2} = 2ga\left(1 + \dfrac{1}{2}\right) \Rightarrow w = \left(\dfrac{7ga}{2}\right)^{\frac{1}{2}}\)	A1

(Corrected from the printed mark scheme: the right-hand side is printed as \(2mga\left(1 + \frac{1}{2}\right)\).)

Alternatives using projectile motion from \(P\)

\(v_P = \left(\dfrac{ga}{2}\right)^{\frac{1}{2}}\), as above
\(\downarrow\) \(u_y = \left(\dfrac{ga}{2}\right)^{\frac{1}{2}}\sin 60^\circ = \left(\dfrac{3ga}{8}\right)^{\frac{1}{2}}\)
\(\downarrow\) \(v_y^2 = u_y^2 + 2g \times \dfrac{3a}{2},\ = \dfrac{27ga}{8}\)	M1, A1
\(\rightarrow\) \(u_x = \left(\dfrac{ga}{2}\right)^{\frac{1}{2}}\cos 60^\circ = \left(\dfrac{ga}{8}\right)^{\frac{1}{2}}\)	A1
\(w^2 = u_x^2 + v_y^2 = \dfrac{ga}{8} + \dfrac{27ga}{8} = \dfrac{7ga}{2} \Rightarrow w = \left(\dfrac{7ga}{2}\right)^{\frac{1}{2}}\)	A1

There are also longer projectile methods using time of flight.

In outline, solving \(\dfrac{3a}{2} = \left(\dfrac{3ga}{8}\right)^{\frac{1}{2}}t + \dfrac{1}{2}gt^2\) gives \(t = \left(\dfrac{3a}{2g}\right)^{\frac{1}{2}}\),
then, using \(v = u + at\) gives \(v_y = \left(\dfrac{3ga}{8}\right)^{\frac{1}{2}} + g\left(\dfrac{3a}{2g}\right)^{\frac{1}{2}} = \left(\dfrac{27ga}{8}\right)^{\frac{1}{2}}\), then as before.	M1 A1