M5 June 2011 Q5

5. A uniform rod \(PQ\), of mass \(m\) and length \(2a\), is made to rotate in a vertical plane with constant angular speed \(\sqrt{\left(\dfrac{g}{a}\right)}\) about a fixed smooth horizontal axis through the end \(P\) of the rod.

Show that, when the rod is inclined at an angle \(\theta\) to the downward vertical, the magnitude of the force exerted on the axis by the rod is \(2mg\left|\cos(\tfrac{1}{2}\theta)\right|\). (8)

Diagram of rod PQ at angle theta to the vertical with force components X and Y at P, accelerations a theta-dot squared and a theta double-dot, and weight mg — Mark scheme

Scheme	Marks
\(\ddot{\theta} = 0\)	B1
\(X - mg\sin\theta = ma\ddot{\theta}\,(= 0)\) \(X = mg\sin\theta\)	M1 A1
\(Y - mg\cos\theta = ma\dot{\theta}^2 = ma\dfrac{g}{a} = mg\) \(Y = mg(1 + \cos\theta)\)	M1 A1
\(R = mg\sqrt{(1 + \cos\theta)^2 + \sin^2\theta}\) \(= mg\sqrt{2(1 + \cos\theta)}\)	M1
\(= mg\sqrt{2.2\cos^2(\tfrac{1}{2}\theta)}\)	DM1
\(= 2mg\left\|\cos(\tfrac{1}{2}\theta)\right\|\) *	A1
	(8 marks)