| Exam Board | Edexcel |
|---|---|
| Module | M5 (Mechanics 5) |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Topic | Circular Motion 1 |
| Difficulty | Challenging +1.8 This M5 question requires calculating the perpendicular force component at a pivot using rotational dynamics. Students must resolve the weight of a composite body (disc + particle), apply the perpendicular distance principle, and use the given moment of inertia with torque equations. While methodical, it demands careful geometric reasoning with the angle β, correct identification of force components, and integration of multiple mechanics concepts—significantly harder than routine M1/M2 problems but standard for Further Maths M5. |
| Spec | 3.03a Force: vector nature and diagrams6.04e Rigid body equilibrium: coplanar forces6.05f Vertical circle: motion including free fall |
A body consists of a uniform plane circular disc, of radius $r$ and mass $2m$, with a particle of mass $3m$ attached to the circumference of the disc at the point $P$. The line $PQ$ is a diameter of the disc. The body is free to rotate in a vertical plane about a fixed smooth horizontal axis, $L$, which is perpendicular to the plane of the disc and passes through $Q$. The body is held with $QP$ making an angle $\beta$ with the downward vertical through $Q$, where $\sin \beta = 0.25$, and released from rest. Find the magnitude of the component, perpendicular to $PQ$, of the force acting on the body at $Q$ at the instant when it is released.
[You may assume that the moment of inertia of the body about $L$ is $15mr^2$.]
[6]
\hfill \mbox{\textit{Edexcel M5 Q4 [6]}}