| Exam Board | OCR MEI |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2007 |
| Session | January |
| Marks | 20 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circular Motion 2 |
| Type | Vertical circle: complete revolution conditions |
| Difficulty | Standard +0.3 This is a standard M3 vertical circle and conical pendulum question with routine application of circular motion principles. Part (a) uses the standard minimum speed condition (T=0 at top) and energy conservation—textbook exercises. Part (b) involves resolving forces on a rotating cone, which is slightly less routine but still a standard M3 topic with straightforward force resolution and friction limits. The calculations are multi-step but follow predictable patterns without requiring novel insight. |
| Spec | 6.05c Horizontal circles: conical pendulum, banked tracks6.05d Variable speed circles: energy methods6.05e Radial/tangential acceleration |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| At highest point, minimum speed when \(T = 0\) | M1 | Considering limiting case |
| \(mg = \frac{mv^2}{r}\), so \(v^2 = gr = 9.8 \times 1.8\) | M1 | Equation of motion at top |
| \(v = \sqrt{17.64} \approx 4.2\ \text{ms}^{-1}\) ✓ | A1 | Correct result shown |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Energy conservation from top to bottom: \(\frac{1}{2}mv_B^2 = \frac{1}{2}mv_T^2 + mg(2r)\) | M1 | Energy equation |
| \(v_B^2 = 17.64 + 4 \times 9.8 \times 1.8 = 17.64 + 70.56 = 88.2\) | A1 | |
| At bottom: \(T - mg = \frac{mv_B^2}{r}\) | M1 | Newton's second law at bottom |
| \(T = m\left(g + \frac{v_B^2}{r}\right) = 5\left(9.8 + \frac{88.2}{1.8}\right)\) | M1 | Substituting values |
| \(T = 5 \times 58.8 = 294\ \text{N}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Resolving vertically: \(N\cos\theta = mg\) | M1 | No friction so only N and weight |
| Resolving horizontally: \(N\sin\theta = m\omega^2 r\) | M1 | Circular motion equation |
| \(\tan\theta = \frac{\omega^2 r}{g} = \frac{8.75^2 \times 0.32}{9.8}\) | M1 | Dividing equations |
| \(= \frac{24.5}{9.8} \times 0.32\)... \(= 0.4\) ✓ | A1 | Correct result shown |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Diagram showing: Normal reaction N (perpendicular to cone surface), Weight \(mg\) downward, Friction F up the slope | B1 | N and mg correct |
| Friction acting up the slope (since \(\omega > 8.75\) tends to make P slip upward) | B1 | Friction direction correct |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\tan\theta = 0.4\), so \(\sin\theta = \frac{1}{\sqrt{1.16}}\), \(\cos\theta = \frac{1}{\sqrt{1.16}} \times \frac{1}{...}\) (or use exact values) | M1 | Finding trig values from \(\tan\theta = 0.4\) |
| Resolving vertically: \(N\cos\theta - F\sin\theta = mg\) | M1 | Vertical resolution with friction |
| Resolving horizontally: \(N\sin\theta + F\cos\theta = m\omega^2 r\) | M1 | Horizontal resolution |
| \(F = \mu N = 0.11N\) | M1 | Using limiting friction |
| Solving to get \(\omega^2 = \frac{g(\tan\theta + \mu)}{r(1 - \mu\tan\theta)}\) or equivalent | M1 | Solving simultaneous equations |
| \(\omega \approx 10.6\ \text{rad s}^{-1}\) | A1 | Correct answer |
# Question 2:
## Part (a)(i)
| Answer | Mark | Guidance |
|--------|------|----------|
| At highest point, minimum speed when $T = 0$ | M1 | Considering limiting case |
| $mg = \frac{mv^2}{r}$, so $v^2 = gr = 9.8 \times 1.8$ | M1 | Equation of motion at top |
| $v = \sqrt{17.64} \approx 4.2\ \text{ms}^{-1}$ ✓ | A1 | Correct result shown |
## Part (a)(ii)
| Answer | Mark | Guidance |
|--------|------|----------|
| Energy conservation from top to bottom: $\frac{1}{2}mv_B^2 = \frac{1}{2}mv_T^2 + mg(2r)$ | M1 | Energy equation |
| $v_B^2 = 17.64 + 4 \times 9.8 \times 1.8 = 17.64 + 70.56 = 88.2$ | A1 | |
| At bottom: $T - mg = \frac{mv_B^2}{r}$ | M1 | Newton's second law at bottom |
| $T = m\left(g + \frac{v_B^2}{r}\right) = 5\left(9.8 + \frac{88.2}{1.8}\right)$ | M1 | Substituting values |
| $T = 5 \times 58.8 = 294\ \text{N}$ | A1 | |
## Part (b)(i)
| Answer | Mark | Guidance |
|--------|------|----------|
| Resolving vertically: $N\cos\theta = mg$ | M1 | No friction so only N and weight |
| Resolving horizontally: $N\sin\theta = m\omega^2 r$ | M1 | Circular motion equation |
| $\tan\theta = \frac{\omega^2 r}{g} = \frac{8.75^2 \times 0.32}{9.8}$ | M1 | Dividing equations |
| $= \frac{24.5}{9.8} \times 0.32$... $= 0.4$ ✓ | A1 | Correct result shown |
## Part (b)(ii)
| Answer | Mark | Guidance |
|--------|------|----------|
| Diagram showing: Normal reaction N (perpendicular to cone surface), Weight $mg$ downward, Friction F up the slope | B1 | N and mg correct |
| Friction acting up the slope (since $\omega > 8.75$ tends to make P slip upward) | B1 | Friction direction correct |
## Part (b)(iii)
| Answer | Mark | Guidance |
|--------|------|----------|
| $\tan\theta = 0.4$, so $\sin\theta = \frac{1}{\sqrt{1.16}}$, $\cos\theta = \frac{1}{\sqrt{1.16}} \times \frac{1}{...}$ (or use exact values) | M1 | Finding trig values from $\tan\theta = 0.4$ |
| Resolving vertically: $N\cos\theta - F\sin\theta = mg$ | M1 | Vertical resolution with friction |
| Resolving horizontally: $N\sin\theta + F\cos\theta = m\omega^2 r$ | M1 | Horizontal resolution |
| $F = \mu N = 0.11N$ | M1 | Using limiting friction |
| Solving to get $\omega^2 = \frac{g(\tan\theta + \mu)}{r(1 - \mu\tan\theta)}$ or equivalent | M1 | Solving simultaneous equations |
| $\omega \approx 10.6\ \text{rad s}^{-1}$ | A1 | Correct answer |
---2
\begin{enumerate}[label=(\alph*)]
\item A light inextensible string has length 1.8 m . One end of the string is attached to a fixed point O , and the other end is attached to a particle of mass 5 kg . The particle moves in a complete vertical circle with centre O , so that the string remains taut throughout the motion. Air resistance may be neglected.
\begin{enumerate}[label=(\roman*)]
\item Show that, at the highest point of the circle, the speed of the particle is at least $4.2 \mathrm {~m} \mathrm {~s} ^ { - 1 }$.
\item Find the least possible tension in the string when the particle is at the lowest point of the circle.
\end{enumerate}\item Fig. 2 shows a hollow cone mounted with its axis of symmetry vertical and its vertex V pointing downwards. The cone rotates about its axis with a constant angular speed of $\omega \mathrm { rad } \mathrm { s } ^ { - 1 }$. A particle P of mass 0.02 kg is in contact with the rough inside surface of the cone, and does not slip. The particle P moves in a horizontal circle of radius 0.32 m . The angle between VP and the vertical is $\theta$.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{b209dbe7-769c-4301-a2f3-108c27c8cefb-3_588_510_1046_772}
\captionsetup{labelformat=empty}
\caption{Fig. 2}
\end{center}
\end{figure}
In the case when $\omega = 8.75$, there is no frictional force acting on P .
\begin{enumerate}[label=(\roman*)]
\item Show that $\tan \theta = 0.4$.
Now consider the case when $\omega$ takes a constant value greater than 8.75.
\item Draw a diagram showing the forces acting on P .
\item You are given that the coefficient of friction between P and the surface is 0.11 . Find the maximum possible value of $\omega$ for which the particle does not slip.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{OCR MEI M3 2007 Q2 [20]}}