A2 June 2019 Q7

EdexcelCurrent spec12 marksVertical Circular Motion

7. A particle, \(P\), of mass \(m\) is attached to one end of a light rod of length \(L\). The other end of the rod is attached to a fixed point \(O\) so that the rod is free to rotate in a vertical plane about \(O\). The particle is held with the rod horizontal and is then projected vertically downwards with speed \(u\). The particle first comes to instantaneous rest at the point \(A\).

(a) Explain why the acceleration of \(P\) at \(A\) is perpendicular to \(OA\). (1)

At the instant when \(P\) is at the point \(A\) the acceleration of \(P\) is in a direction making an angle \(\theta\) with the horizontal. Given that \(u^2 = \dfrac{2gL}{3}\),

(b) find

(i) the magnitude of the acceleration of \(P\) at the point \(A\),

(ii) the size of \(\theta\). (6)

(c) Find, in terms of \(m\) and \(g\), the magnitude of the tension in the rod at the instant when \(P\) is at its lowest point. (5)

Mark scheme (a)

Scheme	Marks	AO
\(v = 0 \Rightarrow \dfrac{v^2}{L} = 0 \Rightarrow\) no acceleration towards \(O\) \(\Rightarrow\) acceleration is perpendicular to \(OA\)	B1	2.4
	(1)

Notes

B1: Clear explanation using \(v = 0\)

P starts at the end of the horizontal rod OP of length L moving down with speed u; at A, above the level of O on the other side, the weight mg acts down and the acceleration acts at angle theta below the horizontal — Mark scheme (b)

Scheme	Marks	AO

Conservation of energy	M1	2.1
\(0 = \dfrac{1}{2}mu^2 - mgL\cos\theta \qquad \left(0 = \dfrac{2gL}{3} - 2gL\cos\theta\right)\)	A1	1.1b
\(\Rightarrow \cos\theta = \dfrac{1}{3}\)	A1	1.1b
Complete strategy to find the angle and \|acceleration\|	M1	3.1a
Magnitude: \(\ g\sin\theta = \dfrac{2\sqrt{2}}{3}g\)	A1	1.1b
\(\theta = 71^\circ\) or better	A1	1.1b
	(6)

Mark scheme (c)

Scheme	Marks	AO
Circular motion	M1	3.1a
\(T - mg = \dfrac{mv^2}{L}\)	A1	1.1b
Energy equation	M1	2.1
\(v^2 = \dfrac{2gL}{3} + 2gL\ \left(= \dfrac{8gL}{3}\right)\)	A1	1.1b
\(T = mg + \dfrac{8mg}{3} = \dfrac{11mg}{3}\)	A1	2.2a
	(5)
	(12 marks)

Notes

M1: Equation for circular motion.
Need all terms and dimensionally correct.
Condone sign errors.

A1: Correct unsimplified equation

M1: Use of conservation of energy.
Require all 3 terms and dimensionally correct.

A1: Correct unsimplified equation

A1: Correct only