| Exam Board | Edexcel |
|---|---|
| Module | M5 (Mechanics 5) |
| Year | 2002 |
| Session | June |
| Marks | 17 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moments of inertia |
| Type | Force at pivot/axis |
| Difficulty | Challenging +1.8 This is a challenging M5 compound pendulum problem requiring parallel axis theorem, energy methods, rotational dynamics, and friction analysis across multiple connected parts. While the techniques are standard for Further Maths mechanics, the multi-step nature, need to coordinate linear and angular motion, and the non-trivial friction condition in part (e) place it well above average difficulty. |
| Spec | 3.03r Friction: concept and vector form3.03t Coefficient of friction: F <= mu*R model6.02i Conservation of energy: mechanical energy principle6.05a Angular velocity: definitions |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Notes |
| \(I = \frac{1}{12}m(4a)^2 + ma^2 = \frac{7ma^2}{3}\) | M1 A1 | (2) |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Notes |
| \(\frac{1}{2}\cdot\frac{7ma^2}{3}\cdot\dot{\theta}^2 = mga\sin\theta\) | M1 A1 | |
| \(\dot{\theta}^2 = \frac{6g\sin\theta}{7a}\) | A1 | (3) |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Notes |
| \(2\dot{\theta}\ddot{\theta} = \frac{6g}{7a}\cos\theta\cdot\dot{\theta} \Rightarrow \ddot{\theta} = \frac{3g\cos\theta}{7a}\) | M1 A1 | (2) |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Notes |
| \(mg\cos\theta - N = ma\ddot{\theta}\) | M1 A1 | |
| \(N = mg\cos\theta - m\cdot\frac{3g\cos\theta}{7}\) | M1 | |
| \(N = \frac{4mg\cos\theta}{7}\) | A1 | (4) |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Notes |
| \(F - mg\sin\theta = ma\dot{\theta}^2\) | M1 A1 | |
| \(F = mg\sin\theta + \frac{mg\cdot 6\sin\theta}{7}\) | M1 | |
| \(F = \frac{13mg\sin\theta}{7}\) | A1 | |
| Slip when \(F = \mu N \Rightarrow \frac{13mg\sin\theta}{7} = \mu\cdot\frac{4mg\cos\theta}{7}\) | M1 | |
| \(\Rightarrow \tan\theta = \frac{4\mu}{13}\) | A1 c.s.o. | (6) |
## Question 6(a):
| Working | Marks | Notes |
|---------|-------|-------|
| $I = \frac{1}{12}m(4a)^2 + ma^2 = \frac{7ma^2}{3}$ | M1 A1 | **(2)** |
---
## Question 6(b):
| Working | Marks | Notes |
|---------|-------|-------|
| $\frac{1}{2}\cdot\frac{7ma^2}{3}\cdot\dot{\theta}^2 = mga\sin\theta$ | M1 A1 | |
| $\dot{\theta}^2 = \frac{6g\sin\theta}{7a}$ | A1 | **(3)** |
---
## Question 6(c):
| Working | Marks | Notes |
|---------|-------|-------|
| $2\dot{\theta}\ddot{\theta} = \frac{6g}{7a}\cos\theta\cdot\dot{\theta} \Rightarrow \ddot{\theta} = \frac{3g\cos\theta}{7a}$ | M1 A1 | **(2)** |
---
## Question 6(d):
| Working | Marks | Notes |
|---------|-------|-------|
| $mg\cos\theta - N = ma\ddot{\theta}$ | M1 A1 | |
| $N = mg\cos\theta - m\cdot\frac{3g\cos\theta}{7}$ | M1 | |
| $N = \frac{4mg\cos\theta}{7}$ | A1 | **(4)** |
---
## Question 6(e):
| Working | Marks | Notes |
|---------|-------|-------|
| $F - mg\sin\theta = ma\dot{\theta}^2$ | M1 A1 | |
| $F = mg\sin\theta + \frac{mg\cdot 6\sin\theta}{7}$ | M1 | |
| $F = \frac{13mg\sin\theta}{7}$ | A1 | |
| Slip when $F = \mu N \Rightarrow \frac{13mg\sin\theta}{7} = \mu\cdot\frac{4mg\cos\theta}{7}$ | M1 | |
| $\Rightarrow \tan\theta = \frac{4\mu}{13}$ | A1 c.s.o. | **(6)** |
---\begin{enumerate}[label=(\alph*)]
\item Show that the moment of inertia of the rod about the edge of the table is $\frac { 7 } { 3 } m a ^ { 2 }$.
The rod is released from rest and rotates about the edge of the table. When the rod has turned through an angle $\theta$, its angular speed is $\dot { \theta }$. Assuming that the rod has not started to slip,
\item show that $\dot { \theta } ^ { 2 } = \frac { 6 g \sin \theta } { 7 a }$,
\item find the angular acceleration of the rod,
\item find the normal reaction of the table on the rod.
The coefficient of friction between the rod and the edge of the table is $\mu$.
\item Show that the rod starts to slip when $\tan \theta = \frac { 4 } { 13 } \mu$\\
(6)
\end{enumerate}
\hfill \mbox{\textit{Edexcel M5 2002 Q6 [17]}}