| Exam Board | OCR MEI |
|---|---|
| Module | Further Mechanics B AS (Further Mechanics B AS) |
| Year | 2021 |
| Session | November |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circular Motion 1 |
| Type | Conical pendulum – horizontal circle in free space (no surface) |
| Difficulty | Moderate -0.3 This is a standard conical pendulum problem requiring resolution of forces and circular motion equations. While it involves multiple parts including force diagrams, tension calculation, and angular speed, all steps follow routine procedures taught in Further Mechanics. The conceptual questions (c-e) about modeling assumptions are straightforward. Slightly easier than average due to being a textbook application with no novel problem-solving required. |
| Spec | 3.03b Newton's first law: equilibrium3.03c Newton's second law: F=ma one dimension6.05b Circular motion: v=r*omega and a=v^2/r6.05c Horizontal circles: conical pendulum, banked tracks |
| Answer | Marks | Guidance |
|---|---|---|
| 5 | (a) | Correct sketch |
| Answer | Marks |
|---|---|
| ‘correct’ angle to vertical. | Might be combined to 50g |
| Answer | Marks | Guidance |
|---|---|---|
| (b) | Radius of chair’s motion is 2+2.5sin50° | B1 |
| 𝑇cos50° = 40𝑔+10𝑔 | M1 | 3.3 |
| weight wrong | Must be trig term with T; g |
| Answer | Marks | Guidance |
|---|---|---|
| 𝑇 =762 (N) | A1 | 1.1 |
| 𝑇sin50° = 50𝑟𝜔2 | M1 | 3.3 |
| 𝜔2 = 𝑇sin50°÷50𝑟 | M1 | 3.4 |
| Answer | Marks |
|---|---|
| find ω2 | If they find v first, must also |
| Answer | Marks | Guidance |
|---|---|---|
| ω = 1.73 (radians per second) | A1 | 1.1 |
| Answer | Marks | Guidance |
|---|---|---|
| (c) | T will be greater because the man is heavier | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| you do the calculation | B1 | 2.2a |
| Answer | Marks | Guidance |
|---|---|---|
| (d) | Because it would be too complicated to include the | |
| mass of the chain in the calculation | B1 | 3.5b |
| Answer | Marks | Guidance |
|---|---|---|
| (e) | The chair and person are modelled as a point mass | B1 |
| Air resistance is neglected | B1 | 1.1 |
Question 5:
5 | (a) | Correct sketch | B1 | 2.5 | 40g and 10g vertically downwards;
tension in chain, which is at a
‘correct’ angle to vertical. | Might be combined to 50g
[1]
(b) | Radius of chair’s motion is 2+2.5sin50° | B1 | 3.1b | 3.915…
𝑇cos50° = 40𝑔+10𝑔 | M1 | 3.3 | Res vert; allow if sin / cos wrong or
weight wrong | Must be trig term with T; g
must be in weight terms
𝑇 =762 (N) | A1 | 1.1
𝑇sin50° = 50𝑟𝜔2 | M1 | 3.3 | Res hor
𝜔2 = 𝑇sin50°÷50𝑟 | M1 | 3.4 | Using their horizontal equation with
a correct form for acceleration to
find ω2 | If they find v first, must also
attempt to find ω
ω = 1.73 (radians per second) | A1 | 1.1 | 1.727166…
[6]
(c) | T will be greater because the man is heavier | B1 | 2.2a | and the vertical equation shows this | SC B1 for both correct but
no explanations (or jus one
explanation) B0 if incorrect
explanations
ω will be the same because mass cancels out when
you do the calculation | B1 | 2.2a
[2]
(d) | Because it would be too complicated to include the
mass of the chain in the calculation | B1 | 3.5b | oe | Allow “so that the weight of
the chain doesn’t have to be
taken into account”
[1]
(e) | The chair and person are modelled as a point mass | B1 | 3.5a | Accept other valid assumptions.
Air resistance is neglected | B1 | 1.1
[2]
5 On a fairground ride, the centre of a horizontal circular frame is attached to the top of a vertical pole, OP . When the frame and pole rotate, OP remains vertical and the frame remains horizontal.
Chairs of mass 10 kg are attached to the frame by means of chains of length 2.5 m . The chains are modelled as being both light and inextensible.
A side view of the situation when the ride is stationary is shown in Fig. 5. A chain fixed to point A on the circular frame supports a chair. The distance OA is 2 m .
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{37798594-8cb0-48aa-8401-090f09e25dff-5_839_1074_641_240}
\captionsetup{labelformat=empty}
\caption{Fig. 5}
\end{center}
\end{figure}
A child of mass 40 kg sits in a chair and, after a short time, the ride is rotating at a steady angular speed of $\omega$ radians per second, with the chain inclined at an angle of $50 ^ { \circ }$ to the downward vertical. The motion of the child and chair is in a horizontal circle.
\begin{enumerate}[label=(\alph*)]
\item Draw a sketch showing the forces acting on the chair when the ride is moving at this angular speed.
\item - Determine the tension in the chain.
\begin{itemize}
\item Determine the value of $\omega$.
\end{itemize}
On another occasion, a man of mass 90 kg sits in the chair; after a short time, the ride is rotating in a horizontal circle at a steady speed of $\omega$ radians per second, with the chain inclined at the same angle of $50 ^ { \circ }$ to the downward vertical.
\item Without any detailed calculations, explain how your answers to part (b) for the child would compare with those for the man.
\item Explain why the chain is modelled as light.
\item State two other modelling assumptions that were used in answering part (b).
\end{enumerate}
\hfill \mbox{\textit{OCR MEI Further Mechanics B AS 2021 Q5 [12]}}