| Exam Board | OCR |
|---|---|
| Module | Further Mechanics (Further Mechanics) |
| Year | 2018 |
| Session | March |
| Marks | 10 |
| Topic | Circular Motion 2 |
| Type | Conical pendulum (horizontal circle) |
| Difficulty | Standard +0.3 This is a standard conical pendulum problem with elastic string requiring resolution of forces, circular motion equations, and Hooke's law. Part (i) is guided ('show that'), parts (ii-iii) involve straightforward application of standard formulas (T = λx/l and simultaneous equations), and part (iv) is simple trigonometry. While it requires multiple techniques, all are routine applications with no novel insight needed, making it slightly easier than average. |
| Spec | 6.02g Hooke's law: T = k*x or T = lambda*x/l6.02h Elastic PE: 1/2 k x^26.05a Angular velocity: definitions6.05b Circular motion: v=r*omega and a=v^2/r |
| Answer | Marks | Guidance |
|---|---|---|
| \(T \sin \theta = 3.5 \times 3^2 \times r\) | M1 | Resolving tension and using NII |
| \(r = L \sin \theta\) | B1 | |
| \(T \sin \theta = 3.5 \times 3^2 \times L \sin \theta\) oe | M1 | Eliminating r |
| \(T = 31.5L\) | A1 [4] | AG; working must be clear |
| Answer | Marks | Guidance |
|---|---|---|
| \(T = \frac{75(L - 0.8)}{0.8}\) | B1 [1] | Correct use of Hooke's law |
| Answer | Marks | Guidance |
|---|---|---|
| \(31.5L = 93.75(L - 0.8)\) | M1 | Solution of simultaneous equations |
| \(T = 38.0\) | A1 | |
| \(L = 1.20\) | A1 [3] |
| Answer | Marks | Guidance |
|---|---|---|
| \(T \cos \theta = mg\) | M1 | Resolve vertically |
| \(25.3°\) | A1 [2] |
## (i)
$T \sin \theta = 3.5 \times 3^2 \times r$ | M1 | Resolving tension and using NII
$r = L \sin \theta$ | B1 |
$T \sin \theta = 3.5 \times 3^2 \times L \sin \theta$ oe | M1 | Eliminating r
$T = 31.5L$ | A1 [4] | AG; working must be clear | may use $T \cos \phi$ instead or $r = L \cos \phi$
## (ii)
$T = \frac{75(L - 0.8)}{0.8}$ | B1 [1] | Correct use of Hooke's law | $(T = 93.75L - 75)$
## (iii)
$31.5L = 93.75(L - 0.8)$ | M1 | Solution of simultaneous equations
$T = 38.0$ | A1 |
$L = 1.20$ | A1 [3] |
## (iv)
$T \cos \theta = mg$ | M1 | Resolve vertically
$25.3°$ | A1 [2] |
---3 A particle $P$ of mass 3.5 kg is attached to one end of a light elastic string of natural length 0.8 m and modulus of elasticity 75 N . The other end of the string is attached to a fixed point $O$. The particle rotates in a horizontal circle with a constant angular velocity of $3 \mathrm { rad } \mathrm { s } ^ { - 1 }$. The centre of the circle is vertically below $O$. The magnitude of the tension in the string is $T \mathrm {~N}$ and the length of the extended string is $L \mathrm {~m}$ (see diagram).\\
\includegraphics[max width=\textwidth, alt={}, center]{a8c9d007-e67f-4637-9e74-630ba9a91442-3_460_424_447_817}\\
(i) By considering the acceleration of $P$, show that $T = 31.5 L$.\\
(ii) Write down another relationship between $T$ and $L$.\\
(iii) Find the value of $T$ and the value of $L$.\\
(iv) Find the angle that the string makes with the downwards vertical through $O$.
\hfill \mbox{\textit{OCR Further Mechanics 2018 Q3 [10]}}