| Exam Board | OCR |
|---|---|
| Module | M4 (Mechanics 4) |
| Year | 2011 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moments |
| Type | Rotation about fixed axis: angular acceleration and velocity |
| Difficulty | Standard +0.8 This M4 question requires applying the work-energy principle to rotational motion and using the equation of rotational motion (moment = Iα) in two different configurations. While the concepts are standard for Further Maths mechanics, it demands careful consideration of changing gravitational moments and requires students to work with moment of inertia of a rod about an endpoint, making it moderately challenging but still within typical M4 scope. |
| Spec | 6.05a Angular velocity: definitions |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| WD by couple is \(C \times \frac{\pi}{2}\) | B1 | |
| Change in PE is \(5 \times 9.8 \times 0.9\) | B1 | Must clearly be PE (not moment) |
| By conservation of energy, \(C \times \frac{\pi}{2} = 5 \times 9.8 \times 0.9\) | M1 | Equation involving WD and PE |
| Moment of couple is 28.1 Nm (3 sf) | A1 | |
| [4] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(I = \frac{4}{3} \times 5 \times 0.9^2 \; (= 5.4)\) | B1 | Can be earned anywhere in the question |
| \(28.075 = 5.4\alpha\) | M1 | Applying \(C = I\alpha\) |
| Angular acceleration is 5.20 rad s\(^{-2}\) (3 sf) | A1 ft | ft is \(C \div I\) |
| [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(28.075 - 5 \times 9.8 \times 0.9 = 5.4\alpha\) | M1 | Rotational equation of motion (3 terms) *(Allow 1.8 instead of 0.9 etc)* |
| Angular acceleration is \((-) 2.97\) rad s\(^{-2}\) (3 sf) | A1 | |
| [2] |
# Question 3:
## Part (i)
| Answer/Working | Marks | Guidance |
|---|---|---|
| WD by couple is $C \times \frac{\pi}{2}$ | B1 | |
| Change in PE is $5 \times 9.8 \times 0.9$ | B1 | Must clearly be PE (not moment) |
| By conservation of energy, $C \times \frac{\pi}{2} = 5 \times 9.8 \times 0.9$ | M1 | Equation involving WD and PE |
| Moment of couple is 28.1 Nm (3 sf) | A1 | |
| | [4] | |
## Part (ii)(a)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $I = \frac{4}{3} \times 5 \times 0.9^2 \; (= 5.4)$ | B1 | Can be earned anywhere in the question |
| $28.075 = 5.4\alpha$ | M1 | Applying $C = I\alpha$ |
| Angular acceleration is 5.20 rad s$^{-2}$ (3 sf) | A1 ft | ft is $C \div I$ |
| | [3] | |
## Part (ii)(b)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $28.075 - 5 \times 9.8 \times 0.9 = 5.4\alpha$ | M1 | Rotational equation of motion (3 terms) *(Allow 1.8 instead of 0.9 etc)* |
| Angular acceleration is $(-) 2.97$ rad s$^{-2}$ (3 sf) | A1 | |
| | [2] | |
---3 A uniform rod $X Y$, of mass 5 kg and length 1.8 m , is free to rotate in a vertical plane about a fixed horizontal axis through $X$. The rod is at rest with $Y$ vertically below $X$ when a couple of constant moment is applied to the rod. It then rotates, and comes instantaneously to rest when $X Y$ is horizontal.\\
(i) Find the moment of the couple.\\
(ii) Find the angular acceleration of the rod
\begin{enumerate}[label=(\alph*)]
\item immediately after the couple is first applied,
\item when $X Y$ is horizontal.
\end{enumerate}
\hfill \mbox{\textit{OCR M4 2011 Q3 [9]}}