M5 June 2013 Q7
7. A uniform circular disc, of radius \(r\) and mass \(m\), is free to rotate in a vertical plane about a fixed smooth horizontal axis. This axis is perpendicular to the plane of the disc and passes through a point \(A\) on the circumference of the disc. The disc is held with \(AB\) horizontal, where \(AB\) is a diameter of the disc, and released from rest.
of the force exerted on the disc by the axis immediately after the disc is released. (11)
When \(AB\) is vertical the disc is instantaneously brought to rest by a horizontal impulse which acts in the plane of the disc and is applied to the disc at \(B\).
| Scheme | Marks |
|---|---|
| \(I_A = \tfrac{1}{2}mr^2 + mr^2 = \dfrac{3mr^2}{2}\) | M1 A1 |
| \(\rightarrow\) \(X = mr\dot{\theta}^2 = 0\) | M1 A1A1 |
| \(\downarrow\) \(mg - Y = mr\ddot{\theta}\) | M1 A1 |
| \(M(A)\) \(mgr = \dfrac{3mr^2}{2}\ddot{\theta}\) | M1 A1 |
| \(Y = \tfrac{1}{3}mg\) | DM1 A1 |
| (11) |
Notes
First M1 for use of parallel axes rule
First A1 for correct expression
Second M1 for resolving horizontally (usual rules)
Second A1 for a correct equation
Third A1 for 0
Third M1 for resolving vertically (usual rules)
Fourth A1 for a correct equation
Fourth M1 for moments about \(A\) (usual rules)
Fifth A1 for a correct equation
Fifth M1, dependent on previous two M marks, for solving for \(Y\).
A1 for answer
| Scheme | Marks |
|---|---|
| \(\dfrac{1}{2}\dfrac{3mr^2}{2}\omega^2 = mgr\) | M1 A1 |
| \(I.2r = \dfrac{3mr^2}{2}\omega\) | M1 A1 |
| \(I = \dfrac{m}{2}\sqrt{3gr}\) | DM1 A1 |
| (6) | |
| (17 marks) |
Notes
First M1 for energy equation
First A1 for a correct equation
Second M1 for angular impulse-momentum equation
Second A1 for a correct equation
Third M1, dependent on previous M’s, for solving for \(I\)
Third A1 for the answer