| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2016 |
| Session | January |
| Marks | 15 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circular Motion 2 |
| Type | Vertical circle: complete revolution conditions |
| Difficulty | Challenging +1.2 This is a standard M3 vertical circle problem with a rod (allowing tension and thrust). Part (a) requires applying energy conservation from starting point to top of circle with the minimum speed condition (T=0 at top). Part (b) uses the tension ratio at top and bottom with energy conservation. While it involves multiple steps and careful equation manipulation, the techniques are entirely standard for M3 circular motion - students would have practiced very similar problems. The given cos α = 4/5 simplifies calculations. More routine than average A-level but requires competent application of mechanics principles. |
| Spec | 1.08a Fundamental theorem of calculus: integration as reverse of differentiation6.02d Mechanical energy: KE and PE concepts6.02i Conservation of energy: mechanical energy principle6.05d Variable speed circles: energy methods6.05e Radial/tangential acceleration |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| For complete circles there must be a speed at the top (or equivalent statement) | B1 | Statement shown in working |
| \(\dfrac{1}{2}mu^2 > \dfrac{mgl}{5}\) | M1A1 | Use energy (and above statement) to obtain an inequality for \(u^2\) |
| \(u > \sqrt{\dfrac{2gl}{5}}\) | A1 | Deduce the given result (4) |
| ALT: Energy equation including speed at top; state \(v^2 > 0\) and use in energy equation | B1,M1 | A1A1 as above |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| NL2 at bottom: \(T_{\max} - mg = m\dfrac{v^2}{l}\) | M1A1 | Attempt NL2 at bottom; acceleration in either form; correct equation inc \(\dfrac{v^2}{r}\) form |
| Energy to bottom: \(\dfrac{1}{2}mv^2 - \dfrac{1}{2}mu^2 = mgl(1+\cos\alpha) = \dfrac{mg\times 9}{5}\) | M1A1 | Energy equation from \(A\) to lowest point; correct energy equation |
| \(T_{\max} = mg + \dfrac{m}{l}\!\left(\dfrac{18gl}{5} + u^2\right)\) | A1 | Eliminate \(v^2\) to obtain max tension in terms of \(m, g, l, u\) |
| NL2 at top: \(T_{\min} + mg = m\dfrac{v^2}{l}\) | M1A1 | Attempt NL2 at top; correct equation inc \(\dfrac{v^2}{r}\) form |
| \(T_{\min} = \dfrac{m}{l}\!\left(u^2 - \dfrac{2}{5}gl\right) - mg\) | M1A1 | Use energy to top to obtain expression for least tension in terms of \(m, g, l, u\); correct expression |
| \(4\!\left(\dfrac{m}{l}\!\left(u^2 - \dfrac{2}{5}gl\right) - mg\right) = mg + \dfrac{m}{l}\!\left(\dfrac{18gl}{5} + u^2\right)\) | ddddM1 | Connect the two tensions using 4 on either side |
| \(u = \sqrt{\dfrac{51gl}{15}} = \sqrt{\dfrac{17gl}{5}}\) | A1cso | Obtain the given result (11) |
# Question 7:
## Part (a):
| Working/Answer | Mark | Guidance |
|---|---|---|
| For complete circles there must be a speed at the top (or equivalent statement) | B1 | Statement shown in working |
| $\dfrac{1}{2}mu^2 > \dfrac{mgl}{5}$ | M1A1 | Use energy (and above statement) to obtain an inequality for $u^2$ |
| $u > \sqrt{\dfrac{2gl}{5}}$ | A1 | Deduce the given result **(4)** |
**ALT:** Energy equation including speed at top; state $v^2 > 0$ and use in energy equation | B1,M1 | A1A1 as above
## Part (b):
| Working/Answer | Mark | Guidance |
|---|---|---|
| NL2 at bottom: $T_{\max} - mg = m\dfrac{v^2}{l}$ | M1A1 | Attempt NL2 at bottom; acceleration in either form; correct equation inc $\dfrac{v^2}{r}$ form |
| Energy to bottom: $\dfrac{1}{2}mv^2 - \dfrac{1}{2}mu^2 = mgl(1+\cos\alpha) = \dfrac{mg\times 9}{5}$ | M1A1 | Energy equation from $A$ to lowest point; correct energy equation |
| $T_{\max} = mg + \dfrac{m}{l}\!\left(\dfrac{18gl}{5} + u^2\right)$ | A1 | Eliminate $v^2$ to obtain max tension in terms of $m, g, l, u$ |
| NL2 at top: $T_{\min} + mg = m\dfrac{v^2}{l}$ | M1A1 | Attempt NL2 at top; correct equation inc $\dfrac{v^2}{r}$ form |
| $T_{\min} = \dfrac{m}{l}\!\left(u^2 - \dfrac{2}{5}gl\right) - mg$ | M1A1 | Use energy to top to obtain expression for least tension in terms of $m, g, l, u$; correct expression |
| $4\!\left(\dfrac{m}{l}\!\left(u^2 - \dfrac{2}{5}gl\right) - mg\right) = mg + \dfrac{m}{l}\!\left(\dfrac{18gl}{5} + u^2\right)$ | ddddM1 | Connect the two tensions using 4 on either side |
| $u = \sqrt{\dfrac{51gl}{15}} = \sqrt{\dfrac{17gl}{5}}$ | A1cso | Obtain the given result **(11)** |
**[Total: 15]**7.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{ffe0bc72-3136-48d9-9d5b-4a364d134070-11_581_641_262_678}
\captionsetup{labelformat=empty}
\caption{Figure 4}
\end{center}
\end{figure}
A particle of mass $m$ is attached to one end of a light rod of length $l$. The other end of the rod is attached to a fixed point $O$. The rod can turn freely in a vertical plane about a horizontal axis through $O$. The particle is projected with speed $u$ from a point $A$, where $O A$ makes an angle $\alpha$ with the upward vertical through $O$, as shown in Figure 4. The particle moves in complete vertical circles.
Given that $\cos \alpha = \frac { 4 } { 5 }$
\begin{enumerate}[label=(\alph*)]
\item show that $u > \sqrt { \frac { 2 g l } { 5 } }$
As the rod rotates, the least tension in the rod is $T$ and the greatest tension is $4 T$.
\item Show that $u = \sqrt { \frac { 17 } { 5 } g l }$
\begin{center}
\includegraphics[max width=\textwidth, alt={}]{ffe0bc72-3136-48d9-9d5b-4a364d134070-12_2639_1830_121_121}
\end{center}
\end{enumerate}
\hfill \mbox{\textit{Edexcel M3 2016 Q7 [15]}}