| Exam Board | OCR |
|---|---|
| Module | M4 (Mechanics 4) |
| Year | 2012 |
| Session | June |
| Marks | 15 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circular Motion 2 |
| Type | Energy considerations in circular motion |
| Difficulty | Challenging +1.2 This is a standard M4 circular motion problem requiring energy conservation, moment of inertia about a parallel axis, and force resolution. While it involves multiple parts and the parallel axis theorem (I = ½ma² + m(½a)² = ¾ma²), each step follows a predictable pattern: (i) energy method for angular speed, (ii) torque equation for angular acceleration, (iii) resolving forces radially and tangentially, (iv) finding resultant force. The techniques are well-rehearsed for M4 students, making this slightly above average difficulty but not requiring novel insight. |
| Spec | 3.04a Calculate moments: about a point3.04b Equilibrium: zero resultant moment and force6.02d Mechanical energy: KE and PE concepts6.02e Calculate KE and PE: using formulae6.02i Conservation of energy: mechanical energy principle6.04e Rigid body equilibrium: coplanar forces6.05a Angular velocity: definitions6.05b Circular motion: v=r*omega and a=v^2/r |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(I = \frac{1}{2}ma^2 + m(\frac{1}{2}a)^2 \quad (= \frac{3}{4}ma^2)\) | B1 | |
| \(\frac{1}{2}I\omega^2 = mg(\frac{1}{2}a)(1-\cos\theta)\) | M1 | Equation involving KE and PE |
| \(\frac{3}{8}ma^2\omega^2 = \frac{1}{2}mga(1-\cos\theta)\) | A1 | |
| Angular speed is \(\sqrt{\dfrac{4g(1-\cos\theta)}{3a}}\) | A1 | AG Correctly obtained |
| [4] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(mg(\frac{1}{2}a\sin\theta) = I\alpha\) | M1 | Equation of rotational motion — *Or differentiation of energy equation* |
| \(\frac{1}{2}mga\sin\theta = \frac{3}{4}ma^2\alpha\) | ||
| Angular acceleration is \(\dfrac{2g\sin\theta}{3a}\) | A1 | |
| [2] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(mg\cos\theta - R = m(\frac{1}{2}a)\omega^2\) | M1 | For radial acceleration \(r\omega^2\) |
| \(mg\cos\theta - R = \frac{2}{3}mg(1-\cos\theta)\) | A1 | |
| \(R = \frac{1}{3}mg(5\cos\theta - 2)\) | A1 | |
| \(mg\sin\theta - S = m(\frac{1}{2}a)\alpha\) | M1 | For transverse acceleration \(r\alpha\) — *Or \(S(\frac{1}{2}a) = I_C\alpha\) (must use \(I_C\))* |
| \(mg\sin\theta - S = \frac{1}{3}mg\sin\theta\) | A1 | FT if incorrect \(r\) already penalised — *Or \(S(\frac{1}{2}a) = (\frac{1}{2}ma^2)\alpha\)* |
| \(S = \frac{2}{3}mg\sin\theta\) | A1 | |
| [6] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| When \(R = 0\), \(\cos\theta = \dfrac{2}{5}\) | M1 | (\(\theta = 1.16\) rad *or* \(66.4°\)) |
| \(\sin\theta = \dfrac{\sqrt{21}}{5}\), \(\quad S = \dfrac{2}{3}mg\left(\dfrac{\sqrt{21}}{5}\right)\) | M1 | Obtaining a value of \(S\) |
| Force exerted by axis is \(\dfrac{2\sqrt{21}}{15}mg\) | A1 | Accept \(0.611mg\) or \(5.99m\) |
| [3] |
## Question 7:
### Part (i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $I = \frac{1}{2}ma^2 + m(\frac{1}{2}a)^2 \quad (= \frac{3}{4}ma^2)$ | B1 | |
| $\frac{1}{2}I\omega^2 = mg(\frac{1}{2}a)(1-\cos\theta)$ | M1 | Equation involving KE and PE |
| $\frac{3}{8}ma^2\omega^2 = \frac{1}{2}mga(1-\cos\theta)$ | A1 | |
| Angular speed is $\sqrt{\dfrac{4g(1-\cos\theta)}{3a}}$ | A1 | AG Correctly obtained |
| **[4]** | | |
### Part (ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $mg(\frac{1}{2}a\sin\theta) = I\alpha$ | M1 | Equation of rotational motion — *Or differentiation of energy equation* |
| $\frac{1}{2}mga\sin\theta = \frac{3}{4}ma^2\alpha$ | | |
| Angular acceleration is $\dfrac{2g\sin\theta}{3a}$ | A1 | |
| **[2]** | | |
### Part (iii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $mg\cos\theta - R = m(\frac{1}{2}a)\omega^2$ | M1 | For radial acceleration $r\omega^2$ |
| $mg\cos\theta - R = \frac{2}{3}mg(1-\cos\theta)$ | A1 | |
| $R = \frac{1}{3}mg(5\cos\theta - 2)$ | A1 | |
| $mg\sin\theta - S = m(\frac{1}{2}a)\alpha$ | M1 | For transverse acceleration $r\alpha$ — *Or $S(\frac{1}{2}a) = I_C\alpha$ (must use $I_C$)* |
| $mg\sin\theta - S = \frac{1}{3}mg\sin\theta$ | A1 | FT if incorrect $r$ already penalised — *Or $S(\frac{1}{2}a) = (\frac{1}{2}ma^2)\alpha$* |
| $S = \frac{2}{3}mg\sin\theta$ | A1 | |
| **[6]** | | |
### Part (iv):
| Answer | Marks | Guidance |
|--------|-------|----------|
| When $R = 0$, $\cos\theta = \dfrac{2}{5}$ | M1 | ($\theta = 1.16$ rad *or* $66.4°$) |
| $\sin\theta = \dfrac{\sqrt{21}}{5}$, $\quad S = \dfrac{2}{3}mg\left(\dfrac{\sqrt{21}}{5}\right)$ | M1 | Obtaining a value of $S$ |
| Force exerted by axis is $\dfrac{2\sqrt{21}}{15}mg$ | A1 | Accept $0.611mg$ or $5.99m$ |
| **[3]** | | |7\\
\includegraphics[max width=\textwidth, alt={}, center]{ab760a4b-e0ec-4256-838f-ed6c762ff18b-4_783_783_255_641}
A uniform circular disc with centre $C$ has mass $m$ and radius $a$. The disc is free to rotate in a vertical plane about a fixed horizontal axis passing through a point $A$ on the disc, where $A C = \frac { 1 } { 2 } a$. The disc is slightly disturbed from rest in the position with $C$ vertically above $A$. When $A C$ makes an angle $\theta$ with the upward vertical the force exerted by the axis on the disc has components $R$ parallel to $A C$ and $S$ perpendicular to $A C$ (see diagram).\\
(i) Show that the angular speed of the disc is $\sqrt { \frac { 4 g ( 1 - \cos \theta ) } { 3 a } }$.\\
(ii) Find the angular acceleration of the disc, in terms of $a , g$ and $\theta$.\\
(iii) Find $R$ and $S$, in terms of $m , g$ and $\theta$.\\
(iv) Find the magnitude of the force exerted by the axis on the disc at an instant when $R = 0$.
\hfill \mbox{\textit{OCR M4 2012 Q7 [15]}}