| Exam Board | OCR |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2010 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circular Motion 1 |
| Type | Particle on cone surface – with string attached to vertex or fixed point |
| Difficulty | Standard +0.3 This is a standard conical pendulum problem with straightforward resolution of forces. Part (i) requires routine application of Newton's second law in horizontal and vertical directions with given geometry (45° makes the algebra clean), then algebraic manipulation to reach the given result. Part (ii) involves recognizing that loss of contact occurs when R=0 and substituting into the derived formula. While it requires careful handling of the geometry and multiple steps, it follows a well-established template for M2 circular motion problems with no novel insight required. |
| Spec | 3.03e Resolve forces: two dimensions3.03m Equilibrium: sum of resolved forces = 06.05c Horizontal circles: conical pendulum, banked tracks |
| Answer | Marks | Guidance |
|---|---|---|
| \(\cos45° + R\sin45° = mg\) | *M1 A1 | 3 terms |
| \(T\sin45° - R\cos45° = m\sin45°\omega^2\) | *M1 A1 | 3 terms; \(a = r\omega^2\) |
| \(2T = \sqrt{2}mg + m\omega^2\) | Dep*M1 | Method to eliminate \(R\) |
| \(T = \frac{m}{2}(\sqrt{2}g + lo^2)\) | A1 6 | AG www |
| Answer | Marks | Guidance |
|---|---|---|
| \(R = 0\) | B1 | may be implied |
| \(2R = \sqrt{2}mg - m\omega^2\) or \(T\cos45° = mg\) or \(T = m\omega^2\) | B1 | |
| Solve to find \(\omega\) | M1 | |
| \(\omega = 4.16 \text{ rad s}^{-1}\) | A1 4 | 10 |
## (i)
$\cos45° + R\sin45° = mg$ | *M1 A1 | 3 terms
$T\sin45° - R\cos45° = m\sin45°\omega^2$ | *M1 A1 | 3 terms; $a = r\omega^2$
$2T = \sqrt{2}mg + m\omega^2$ | Dep*M1 | Method to eliminate $R$
$T = \frac{m}{2}(\sqrt{2}g + lo^2)$ | A1 6 | AG www
## (ii)
$R = 0$ | B1 | may be implied
$2R = \sqrt{2}mg - m\omega^2$ or $T\cos45° = mg$ or $T = m\omega^2$ | B1 |
Solve to find $\omega$ | M1 |
$\omega = 4.16 \text{ rad s}^{-1}$ | A1 4 | 10
---One end of a light inextensible string of length $l$ is attached to the vertex of a smooth cone of semi-vertical angle $45°$. The cone is fixed to the ground with its axis vertical. The other end of the string is attached to a particle of mass $m$ which rotates in a horizontal circle in contact with the outer surface of the cone. The angular speed of the particle is $\omega$ (see diagram). The tension in the string is $T$ and the contact force between the cone and the particle is $R$.
\begin{enumerate}[label=(\roman*)]
\item By resolving horizontally and vertically, find two equations involving $T$ and $R$ and hence show that $T = \frac{1}{2}ml(\sqrt{2}g + l\omega^2)$. [6]
\item When the string has length 0.8 m, calculate the greatest value of $\omega$ for which the particle remains in contact with the cone. [4]
\end{enumerate}
\hfill \mbox{\textit{OCR M2 2010 Q5 [10]}}