Standard +0.3 This is a standard conical pendulum problem requiring resolution of forces (tension and weight), application of circular motion formula (F=mrω²), and basic trigonometry/Pythagoras. While it's an 8-mark 'show that' question requiring multiple steps, it follows a well-established method taught in M3 with no novel insight needed—slightly easier than average due to its routine nature.
A particle is attached to one end of a light inextensible string of length \(l\). The other end of the string is attached to a fixed point \(A\). The particle moves with constant angular speed \(\omega\) in a horizontal circle. The centre of the circle is vertically below \(A\) and the radius of the circle is \(r\).
Show that \(\omega^2 = \frac{g}{\sqrt{l^2 - r^2}}\) [8]
- Triangle of forces methods must show the triangle (not just the equations) for the working to be complete.
| Answer/Working | Marks | Guidance |
|---|---|---|
| $T \cos \theta = mg$ | M1A1 | Attempting to resolve vertically |
| $T \sin \theta = mr\omega^2$ | M1A1 | Attempting NL2 along the radius, acceleration in either form |
| $\tan \theta = \frac{r\omega^2}{g}$ | DM1 | Eliminate $T$ Dependent on 1st and 2nd M marks |
| $\tan \theta = \frac{r}{\sqrt{l^2 - r^2}}$ | B1 | $\tan \theta$ or $\cos \theta$ (ie vert side of triangle used) |
| $\frac{r\omega^2}{g} = \frac{r}{\sqrt{l^2 - r^2}}$ | DM1 | Eliminate the angle Dependent on 1st and 2nd M marks |
| $\omega^2 = \frac{g}{\sqrt{l^2 - r^2}}$ * | A1cso | Obtain the GIVEN result |
**Total: [8]**
**Additional Notes:**
- $mg \sin \theta = mr\omega^2 \cos \theta$ as initial equation scores M2A2
- Triangle of forces methods must show the triangle (not just the equations) for the working to be complete.
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A particle is attached to one end of a light inextensible string of length $l$. The other end of the string is attached to a fixed point $A$. The particle moves with constant angular speed $\omega$ in a horizontal circle. The centre of the circle is vertically below $A$ and the radius of the circle is $r$.
Show that $\omega^2 = \frac{g}{\sqrt{l^2 - r^2}}$ [8]
\hfill \mbox{\textit{Edexcel M3 2016 Q1 [8]}}