Challenging +1.2 This is a multi-step vertical circle problem requiring energy conservation and force analysis at an unknown position. Students must find the speed at the lowest point from the maximum tension condition, then use energy methods to find the angle where the given speed occurs. While it involves several standard techniques (tension formula, energy conservation, solving a trigonometric equation), the problem is straightforward for M3 students who have practiced vertical circle problems with rods.
3. A light rod \(A B\) of length \(2 a\) has a particle \(P\) of mass \(m\) attached to \(B\). The rod is rotating in a vertical plane about a fixed smooth horizontal axis through \(A\). Given that the greatest tension in the rod is \(\frac { 9 m g } { 8 }\), find, to the nearest degree, the angle between the rod and the downward vertical when the speed of \(P\) is \(\sqrt { \left( \frac { a g } { 20 } \right) }\).
M1: energy equation from bottom or top to point where speed is \(\sqrt{\frac{ag}{20}}\). Must have difference of KE terms and a GPE term. A1ft: correct difference of KE terms or correct PE term (follow through their \(u\)). A1: completely correct equation
\(\theta = 18°\) nearest degree
M1dep A1
M1dep: substituting \(v = \sqrt{\frac{ag}{20}}\) and solving for \(\theta\), dependent on both previous M marks. A1cao: must be nearest degree
## Question 3:
| Working/Answer | Marks | Guidance |
|---|---|---|
| $\frac{9}{8}mg - mg = \frac{mu^2}{2a}$ | M1 A1 | M1: NL2 along radius at bottom or top. Must have 2 forces and an acceleration. A1: fully correct equation, must be at the bottom |
| $u^2 = \frac{ag}{4}$ | A1 | |
| $\frac{1}{2}m\left(\frac{ag}{4}\right) - \frac{1}{2}m\left(\frac{ag}{20}\right) = mg \cdot 2a(1-\cos\theta)$ | M1 A1 A1 | M1: energy equation from bottom or top to point where speed is $\sqrt{\frac{ag}{20}}$. Must have difference of KE terms and a GPE term. A1ft: correct difference of KE terms or correct PE term (follow through their $u$). A1: completely correct equation |
| $\theta = 18°$ nearest degree | M1dep A1 | M1dep: substituting $v = \sqrt{\frac{ag}{20}}$ and solving for $\theta$, dependent on both previous M marks. A1cao: must be nearest degree |
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3. A light rod $A B$ of length $2 a$ has a particle $P$ of mass $m$ attached to $B$. The rod is rotating in a vertical plane about a fixed smooth horizontal axis through $A$. Given that the greatest tension in the rod is $\frac { 9 m g } { 8 }$, find, to the nearest degree, the angle between the rod and the downward vertical when the speed of $P$ is $\sqrt { \left( \frac { a g } { 20 } \right) }$.\\
\hfill \mbox{\textit{Edexcel M3 2014 Q3 [8]}}