| Exam Board | OCR |
|---|---|
| Module | FM1 AS (Further Mechanics 1 AS) |
| Year | 2017 |
| Session | December |
| Marks | 10 |
| Topic | Circular Motion 1 |
| Type | Particle on table with string above |
| Difficulty | Standard +0.3 This is a straightforward circular motion problem requiring standard formulas (v=rω, resolving forces vertically, finding limiting case when N=0). Part (i) is verification using v=rω, part (ii) requires resolving T vertically against weight and normal force, and part (iii) applies the limiting condition. All steps are routine applications of FM1 circular motion principles with no novel insight required, making it slightly easier than average for Further Maths. |
| Spec | 6.05b Circular motion: v=r*omega and a=v^2/r6.05c Horizontal circles: conical pendulum, banked tracks |
| Answer | Marks | Guidance |
|---|---|---|
| \(r = \frac{0.5}{0.625} = 0.8\) | B1 [1] | AG; use of \(v = r\omega\) must be indicated |
| Answer | Marks | Guidance |
|---|---|---|
| \(\tan\theta = \frac{0.8}{1.8}\) oe | B1 | soi, eg by correct value of an angle or its sine or cosine |
| Answer | Marks | Guidance |
|---|---|---|
| \(T\cos\theta + R = mg\) | M1, A1 | For attempt at either resolution (3 forces required for vertical equation; Newton II for horizontal equation) |
| \(T\sin\theta = \frac{mv^2}{r}\) or \(T\sin\theta = m\omega^2 r\) | A1 | |
| Solve pair of equations to find R \(R = 1.82\) | M1, A1 [6] | At any stage; Equations must include attempt to resolve T; Candidates may work with angle to the horizontal instead |
| Answer | Marks | Guidance |
|---|---|---|
| \(T\cos\theta = mg\) and \(T\sin\theta = \frac{mv^2}{r}\) | B1 | Two equations, with condition \(R = 0\) soi |
| \(\tan\theta = \frac{v^2}{rg}\) | A1 | oe |
| \(v = 1.87\) | A1 [3] |
## (i)
$r = \frac{0.5}{0.625} = 0.8$ | B1 [1] | AG; use of $v = r\omega$ must be indicated
## (ii)
$\tan\theta = \frac{0.8}{1.8}$ oe | B1 | soi, eg by correct value of an angle or its sine or cosine
**Resolving vertically and horizontally:**
$T\cos\theta + R = mg$ | M1, A1 | For attempt at either resolution (3 forces required for vertical equation; Newton II for horizontal equation)
$T\sin\theta = \frac{mv^2}{r}$ or $T\sin\theta = m\omega^2 r$ | A1 |
**Solve pair of equations to find R** $R = 1.82$ | M1, A1 [6] | At any stage; Equations must include attempt to resolve T; Candidates may work with angle to the horizontal instead
## (iii)
$T\cos\theta = mg$ and $T\sin\theta = \frac{mv^2}{r}$ | B1 | Two equations, with condition $R = 0$ soi
$\tan\theta = \frac{v^2}{rg}$ | A1 | oe
$v = 1.87$ | A1 [3] |
---6\\
\includegraphics[max width=\textwidth, alt={}, center]{a1a43547-0a68-4346-884a-0c6d9302cf24-4_547_597_251_735}
A particle of mass 0.2 kg is attached to one end of a light inextensible string. The other end of the string is attached to a fixed point $O$ which is 1.8 m above a smooth horizontal table. The particle moves on the table in a circular path at constant speed with the string taut (see diagram).
The particle has a speed of $0.5 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ and its angular velocity is $0.625 \mathrm { rad } \mathrm { s } ^ { - 1 }$.\\
(i) Show that the radius of the circular path is 0.8 m .\\
(ii) Find the magnitude of the normal contact force between the particle and the table.
The speed is changed to $v \mathrm {~ms} ^ { - 1 }$. At this speed the particle is just about to lose contact with the table.\\
(iii) Find the value of $v$.
\hfill \mbox{\textit{OCR FM1 AS 2017 Q6 [10]}}