| Exam Board | Edexcel |
|---|---|
| Module | M5 (Mechanics 5) |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Topic | Circular Motion 1 |
| Type | Centre of mass of rotating body |
| Difficulty | Standard +0.8 This M5 question requires setting up equations of motion for a rotating pulley system with hanging masses, applying Newton's second law to both particles and the rotational equation to the pulley, then solving simultaneously for angular acceleration. Part (b) adds rotational kinematics with a braking couple. While systematic, it demands careful handling of multiple connected components, tension differences, and the moment of inertia of a disc (I = 2mr²), making it moderately challenging but still a standard mechanics problem type. |
| Spec | 3.03k Connected particles: pulleys and equilibrium6.05a Angular velocity: definitions6.05b Circular motion: v=r*omega and a=v^2/r |
A uniform circular pulley, of mass $4m$ and radius $r$, is free to rotate about a fixed smooth horizontal axis which passes through the centre of the pulley and is perpendicular to the plane of the pulley. A light inextensible string passes over the pulley and has a particle of mass $2m$ attached to one end and a particle of mass $3m$ attached to the other end. The particles hang with the string vertical and taut on each side of the pulley. The rim of the pulley is sufficiently rough to prevent the string slipping. The system is released from rest.
\begin{enumerate}[label=(\alph*)]
\item Find the angular acceleration of the pulley.
[8]
\end{enumerate}
When the angular speed of the pulley is $\Omega$, the string breaks and a constant braking couple of magnitude $G$ is applied to the pulley which brings it to rest.
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item Find an expression for the angle turned through by the pulley from the instant when the string breaks to the instant when the pulley first comes to rest.
[4]
\end{enumerate}
\hfill \mbox{\textit{Edexcel M5 Q6 [12]}}