A2 June 2024 Q3

Figure 2 shows a hemispherical bowl of internal radius \(10d\) that is fixed with its circular rim horizontal.

The centre of the circular rim is at the point \(O\).

A particle \(P\) moves with constant angular speed on the smooth inner surface of the bowl.

The particle \(P\) moves in a horizontal circle with radius \(8d\) and centre \(C\).

(a) Find, in terms of \(g\), the exact magnitude of the acceleration of \(P\). (6)

The time for \(P\) to complete one revolution is \(T\).

(b) Find \(T\) in terms of \(d\) and \(g\). (3)

Scheme	Marks	AO

Resolve vertically	M1	3.3
\(R\sin\theta = mg\)	A1	1.1b
Horizontal equation of motion	M1	3.3
\(R\cos\theta = ma\ \left(= mr\omega^2\right)\)	A1	1.1b
Solve for \(a\)	DM1	2.1
\(\dfrac{g}{a} = \tan\theta \Rightarrow a = \dfrac{4}{3}g\)	A1	1.1b
	(6)

Mark scheme (b)

Scheme	Marks	AO
\(8d\omega^2 = \dfrac{4}{3}g \Rightarrow \omega = \sqrt{\dfrac{g}{6d}}\)	M1	3.4
Use of \(T = \dfrac{2\pi}{\omega}\)	M1	1.1b
\(T = 2\pi\sqrt{\dfrac{6d}{g}}\) oe OR \(15\sqrt{\dfrac{d}{g}}\) or better	A1	1.1b
	(3)
	(9 marks)

M1: Use \(a = r\omega^2\) or \(a = \dfrac{v^2}{r}\) to obtain \(\omega\) or \(v\) \(\left(v = \sqrt{\dfrac{32dg}{3}}\right)\)

M1: Complete method to find \(T\)

A1: \(15\sqrt{\dfrac{d}{g}}\) or better - must be in terms of \(d\) and \(g\). Accept \(T = \dfrac{2\pi}{\sqrt{\dfrac{g}{6d}}}\)