| Exam Board | OCR MEI |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2006 |
| Session | June |
| Marks | 18 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circular Motion 1 |
| Type | Conical pendulum – horizontal circle in free space (no surface) |
| Difficulty | Standard +0.3 This is a straightforward application of standard circular motion principles. Part (a) involves resolving forces for a conical pendulum using basic trigonometry and F=mrω². Part (b) requires finding friction components (one for angular acceleration, one for centripetal force) and applying the friction limit F=μR. All steps are routine M3 techniques with no novel problem-solving required, making it slightly easier than average. |
| Spec | 3.03d Newton's second law: 2D vectors6.05a Angular velocity: definitions6.05b Circular motion: v=r*omega and a=v^2/r6.05c Horizontal circles: conical pendulum, banked tracks |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Resolving vertically: \(T\cos 55° = mg\) | M1 | |
| \(T = \dfrac{0.6 \times 9.8}{\cos 55°} = 10.2\) N | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(r = 2.8\sin 55°\) | M1 | |
| \(T\sin 55° = \dfrac{mv^2}{r}\) | M1 | |
| \(v^2 = \dfrac{T\sin 55° \times r}{m} = \dfrac{T\sin^2 55° \times 2.8}{0.6}\) | M1 | |
| \(v = \sqrt{\dfrac{10.2 \times \sin^2 55° \times 2.8}{0.6}}\) | A1 | \(v \approx 5.73\) ms\(^{-1}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Tangential (perpendicular to QO): \(F_1 = mr\alpha = 0.5 \times 1.4 \times 1.12\) | M1 | |
| \(F_1 = 0.784\) N | A1 | |
| Radial (parallel to QO, towards O): \(F_2 = mr\omega^2 = 0.5 \times 1.4\omega^2\) | M1 | |
| \(F_2 = 0.7\omega^2\) N | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(R = mg = 0.5 \times 9.8 = 4.9\) N | B1 | |
| Limiting friction \(= 0.65 \times 4.9 = 3.185\) N | M1 | |
| Total friction \(F = \sqrt{F_1^2 + F_2^2} = \sqrt{0.784^2 + 0.49\omega^4}\) | M1 A1 | |
| \(0.784^2 + 0.49\omega^4 = 3.185^2\) | M1 | |
| \(\omega^4 = \dfrac{3.185^2 - 0.784^2}{0.49}\) | A1 | |
| \(\omega \approx 2.40\) rad s\(^{-1}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\tan\theta = \dfrac{F_1}{F_2} = \dfrac{0.784}{0.7\omega^2}\) | M1 | |
| Substituting \(\omega\): \(\theta \approx 15.1°\) | A1 A1 |
# Question 2:
## Part (a)(i) - Tension in string
| Answer | Mark | Guidance |
|--------|------|----------|
| Resolving vertically: $T\cos 55° = mg$ | M1 | |
| $T = \dfrac{0.6 \times 9.8}{\cos 55°} = 10.2$ N | A1 | |
## Part (a)(ii) - Speed of P
| Answer | Mark | Guidance |
|--------|------|----------|
| $r = 2.8\sin 55°$ | M1 | |
| $T\sin 55° = \dfrac{mv^2}{r}$ | M1 | |
| $v^2 = \dfrac{T\sin 55° \times r}{m} = \dfrac{T\sin^2 55° \times 2.8}{0.6}$ | M1 | |
| $v = \sqrt{\dfrac{10.2 \times \sin^2 55° \times 2.8}{0.6}}$ | A1 | $v \approx 5.73$ ms$^{-1}$ |
## Part (b)(i) - Components $F_1$ and $F_2$
| Answer | Mark | Guidance |
|--------|------|----------|
| Tangential (perpendicular to QO): $F_1 = mr\alpha = 0.5 \times 1.4 \times 1.12$ | M1 | |
| $F_1 = 0.784$ N | A1 | |
| Radial (parallel to QO, towards O): $F_2 = mr\omega^2 = 0.5 \times 1.4\omega^2$ | M1 | |
| $F_2 = 0.7\omega^2$ N | A1 | |
## Part (b)(ii) - Value of $\omega$ when about to slip
| Answer | Mark | Guidance |
|--------|------|----------|
| $R = mg = 0.5 \times 9.8 = 4.9$ N | B1 | |
| Limiting friction $= 0.65 \times 4.9 = 3.185$ N | M1 | |
| Total friction $F = \sqrt{F_1^2 + F_2^2} = \sqrt{0.784^2 + 0.49\omega^4}$ | M1 A1 | |
| $0.784^2 + 0.49\omega^4 = 3.185^2$ | M1 | |
| $\omega^4 = \dfrac{3.185^2 - 0.784^2}{0.49}$ | A1 | |
| $\omega \approx 2.40$ rad s$^{-1}$ | A1 | |
## Part (b)(iii) - Angle friction makes with QO
| Answer | Mark | Guidance |
|--------|------|----------|
| $\tan\theta = \dfrac{F_1}{F_2} = \dfrac{0.784}{0.7\omega^2}$ | M1 | |
| Substituting $\omega$: $\theta \approx 15.1°$ | A1 A1 | |
---2
\begin{enumerate}[label=(\alph*)]
\item A particle P of mass 0.6 kg is connected to a fixed point by a light inextensible string of length 2.8 m . The particle P moves in a horizontal circle as a conical pendulum, with the string making a constant angle of $55 ^ { \circ }$ with the vertical.
\begin{enumerate}[label=(\roman*)]
\item Find the tension in the string.
\item Find the speed of P .
\end{enumerate}\item A turntable has a rough horizontal surface, and it can rotate about a vertical axis through its centre O . While the turntable is stationary, a small object Q of mass 0.5 kg is placed on the turntable at a distance of 1.4 m from O . The turntable then begins to rotate, with a constant angular acceleration of $1.12 \mathrm { rad } \mathrm { s } ^ { - 2 }$. Let $\omega \mathrm { rad } \mathrm { s } ^ { - 1 }$ be the angular speed of the turntable.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{5bb02383-91c0-4454-aaea-0bd6af6ba325-3_517_522_870_769}
\captionsetup{labelformat=empty}
\caption{Fig. 2}
\end{center}
\end{figure}
\begin{enumerate}[label=(\roman*)]
\item Given that Q does not slip, find the components $F _ { 1 }$ and $F _ { 2 }$ of the frictional force acting on Q perpendicular and parallel to QO (see Fig. 2). Give your answers in terms of $\omega$ where appropriate.
The coefficient of friction between Q and the turntable is 0.65 .
\item Find the value of $\omega$ when Q is about to slip.
\item Find the angle which the frictional force makes with QO when Q is about to slip.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{OCR MEI M3 2006 Q2 [18]}}