M5 June 2017 Q5

EdexcelOld spec15 marksFurther Dynamics Vertical Circular Motion

5. A uniform rod \(AB\), of mass \(M\) and length \(2L\), is free to rotate in a vertical plane about a smooth fixed horizontal axis through \(A\). The rod is hanging vertically at rest, with \(B\) below \(A\), when it is struck at its midpoint by a particle of mass \(\dfrac{1}{2}M\). Immediately before this impact, the particle is moving with speed \(u\), in a direction which is horizontal and perpendicular to the axis. The particle is brought to rest by the impact and immediately after the impact the rod moves with angular speed \(\omega\).

(a) Show that \(\omega = \dfrac{3u}{8L}\) (4)

Immediately after the impact, the magnitude of the vertical component of the force exerted on the rod at \(A\) by the axis is \(\dfrac{3Mg}{2}\)

(b) Find \(u\) in terms of \(L\) and \(g\). (4)

(c) Show that the magnitude of the horizontal component of the force exerted on the rod at \(A\) by the axis, immediately after the impact, is zero. (2)

The rod first comes to instantaneous rest after it has turned through an angle \(\alpha\).

(d) Find the size of \(\alpha\). (5)

Mark scheme (a)

Scheme	Marks
\(\dfrac{1}{2}MuL = \dfrac{1}{3}M(2L)^2\omega\)	M1 A2
\(\omega = \dfrac{3u}{8L}\) PRINTED ANSWER	A1
	(4)

Notes

M1 for conservation of angular momentum with usual rules

First A1 and second A1 for each side

Third A1 for the answer

Mark scheme (b)

Scheme	Marks
\(\dfrac{3Mg}{2} - Mg = ML\omega^2\)	M1 A1
\(\dfrac{Mg}{2} = ML\left(\dfrac{3u}{8L}\right)^2\)	DM1
\(u = \sqrt{\dfrac{32gL}{9}}\)	A1
	(4)

Notes

First M1 for resolving inwards

First A1 for a correct equation

Second DM1, dependent on previous M, for substituting for \(\omega\)

Second A1 for the answer

Mark scheme (c)

Scheme	Marks
M(\(A\)) \(0 = I\ddot{\theta} \Rightarrow \ddot{\theta} = 0\)	M1
\((\rightarrow)\) \(X = ML\ddot{\theta} = 0\)	A1
	(2)

Notes

M1 for moments equn to show that \(\ddot{\theta} = 0\)

A1 for resolving horizontally to give zero.

Mark scheme (d)

Scheme	Marks
\(\dfrac{1}{2}\left(\dfrac{4ML^2}{3}\right)\omega^2 = MgL(1 - \cos\alpha)\)	M1 A1 A1
\(\dfrac{1}{3} = (1 - \cos\alpha)\)	DM1
\(\alpha = \arccos\left(\dfrac{2}{3}\right) = 48^\circ\) or better	A1
	(5)
	(15 marks)

Notes

First M1 for energy equation

First A1 for KE in terms of \(M\), \(L\) and \(\omega\)

Second A1 for PE

Second DM1, dependent on 1^st M, for sub for \(\omega\) and \(u\) to give equn in \(\alpha\) only

Third A1 for 48\(^\circ\) or better (0.84 rad or better)