A2 June 2019 Q1

EdexcelCurrent spec6 marksHorizontal Circular Motion

Figure 1: hemispherical shell of radius a, rim horizontal and uppermost; bead B moves in a horizontal circle whose centre is a/4 below the level of the rim — Figure 1

A hemispherical shell of radius \(a\) is fixed with its rim uppermost and horizontal. A small bead, \(B\), is moving with constant angular speed, \(\omega\), in a horizontal circle on the smooth inner surface of the shell. The centre of the path of \(B\) is at a distance \(\dfrac{1}{4}a\) vertically below the level of the rim of the hemisphere, as shown in Figure 1.

Find the magnitude of \(\omega\), giving your answer in terms of \(a\) and \(g\). (6)

cross-section: reaction R along the radius from B to the centre of the rim at angle theta to the vertical; centre of the circle a/4 below the rim; radius r; weight mg — Mark scheme

Scheme	Marks	AO

\(\updownarrow R\cos\theta = mg\)	M1	3.1b
\(\leftrightarrow R\sin\theta = mr\omega^2\)	M1	3.3
	A1	1.1b
\(\tan\theta = \dfrac{r}{a/4} \qquad \left(\tan\theta = \sqrt{15}\right)\)	B1	1.1b
Complete strategy to find \(\omega\)	M1	3.1b
\(\tan\theta = \dfrac{mr\omega^2}{mg} = \dfrac{4r}{a},\quad \Rightarrow \omega^2 = \dfrac{4g}{a},\quad \omega = 2\sqrt{\dfrac{g}{a}}\)	A1	1.1b
	(6)
	(6 marks)

Scheme	Marks	AO
\(\updownarrow R\cos\theta = mg\)	M1	3.1b
\(\leftrightarrow R\sin\theta = ma\sin\theta\,\omega^2\)	M1	3.3
	A1	1.1b
\(\cos\theta = \tfrac{1}{4}\)	B1	1.1b
Complete strategy to find \(\omega\)	M1	3.1b
\(\Rightarrow R = 4mg,\quad 4mg = ma\omega^2 \quad\Rightarrow\quad \omega = 2\sqrt{\dfrac{g}{a}}\)	A1	1.1b

A2 June 2019 Q1

A2 June 2019 Q1

Notes

Alternative