| Exam Board | OCR |
|---|---|
| Module | M4 (Mechanics 4) |
| Year | 2013 |
| Session | June |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moments of inertia |
| Type | Force at pivot/axis |
| Difficulty | Challenging +1.8 This is a challenging M4 rotation dynamics problem requiring moment of inertia calculation via perpendicular axis theorem, energy conservation for angular speed, torque equation for angular acceleration, and resolution of forces with centripetal/tangential components. It demands multiple advanced techniques and careful geometric reasoning, but follows a standard M4 framework without 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 | Mark | Guidance |
| \(I = \frac{1}{3}m\left[(3a)^2+(4a)^2\right] + m(5a)^2\) | B1 | For \(I_G = \frac{1}{3}m\left[(3a)^2+(4a)^2\right]\) |
| M1 | For \(I_G + m(AG)^2\) | |
| OR \(I = \frac{4}{3}m(3a)^2 + \frac{4}{3}m(4a)^2\) | B1 For \(I_{AD}=\frac{4}{3}m(3a)^2\), \(I_{AB}=\frac{4}{3}m(4a)^2\) \ | |
| \(I = \frac{100}{3}ma^2\) | A1 | |
| [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\frac{1}{2}I\omega^2 = mg(4a+5a\sin\theta)\) | M1 | Equation involving KE and PE |
| \(\frac{50}{3}ma^2\omega^2 = mga(4+5\sin\theta)\) | A1 | FT |
| Angular speed \(\omega = \sqrt{\frac{3g(4+5\sin\theta)}{50a}}\) | A1 | AG |
| [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(mg(5a\cos\theta) = I\alpha\) | M1 | Equation of rotational motion \ |
| Angular acceleration is \(\frac{3g\cos\theta}{20a}\) | A1 | Accept \(\frac{15g\cos\theta}{100a}\) etc |
| [2] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(R - mg\sin\theta = m(5a)\omega^2\) | M1 | For radial acceleration \(r\omega^2\) |
| \(R - mg\sin\theta = \frac{3}{10}mg(4+5\sin\theta)\) | A1 | |
| \(R = \frac{1}{10}mg(12+25\sin\theta)\) | A1 | |
| M1 | For transverse acceleration \(r\alpha\) \ | Or use of \(I_G\alpha\) |
| \(mg\cos\theta - S = m(5a)\alpha\) | A1 | Or \(S(5a) = (\frac{25}{3}ma^2)\alpha\) |
| \(mg\cos\theta - S = \frac{3}{4}mg\cos\theta\) | ||
| \(S = \frac{1}{4}mg\cos\theta\) | A1 | |
| [6] |
## Question 7(i):
| Answer | Mark | Guidance |
|--------|------|----------|
| $I = \frac{1}{3}m\left[(3a)^2+(4a)^2\right] + m(5a)^2$ | B1 | For $I_G = \frac{1}{3}m\left[(3a)^2+(4a)^2\right]$ |
| | M1 | For $I_G + m(AG)^2$ |
| **OR** $I = \frac{4}{3}m(3a)^2 + \frac{4}{3}m(4a)^2$ | | B1 For $I_{AD}=\frac{4}{3}m(3a)^2$, $I_{AB}=\frac{4}{3}m(4a)^2$ \| M1 For $I_{AD}+I_{AB}$ |
| $I = \frac{100}{3}ma^2$ | A1 | |
| **[3]** | | |
## Question 7(ii):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\frac{1}{2}I\omega^2 = mg(4a+5a\sin\theta)$ | M1 | Equation involving KE and PE |
| $\frac{50}{3}ma^2\omega^2 = mga(4+5\sin\theta)$ | A1 | FT |
| Angular speed $\omega = \sqrt{\frac{3g(4+5\sin\theta)}{50a}}$ | A1 | AG |
| **[3]** | | |
## Question 7(iii):
| Answer | Mark | Guidance |
|--------|------|----------|
| $mg(5a\cos\theta) = I\alpha$ | M1 | Equation of rotational motion \| Or differentiating energy equation \| Or writing $\omega\frac{d\omega}{d\theta}$ in terms of $\theta$ |
| Angular acceleration is $\frac{3g\cos\theta}{20a}$ | A1 | Accept $\frac{15g\cos\theta}{100a}$ etc |
| **[2]** | | |
## Question 7(iv):
| Answer | Mark | Guidance |
|--------|------|----------|
| $R - mg\sin\theta = m(5a)\omega^2$ | M1 | For radial acceleration $r\omega^2$ |
| $R - mg\sin\theta = \frac{3}{10}mg(4+5\sin\theta)$ | A1 | |
| $R = \frac{1}{10}mg(12+25\sin\theta)$ | A1 | |
| | M1 | For transverse acceleration $r\alpha$ \| Or use of $I_G\alpha$ |
| $mg\cos\theta - S = m(5a)\alpha$ | A1 | Or $S(5a) = (\frac{25}{3}ma^2)\alpha$ |
| $mg\cos\theta - S = \frac{3}{4}mg\cos\theta$ | | |
| $S = \frac{1}{4}mg\cos\theta$ | A1 | |
| **[6]** | | |7\\
\includegraphics[max width=\textwidth, alt={}, center]{6e3d5f5e-7ffa-4111-903d-468fb4d20192-5_584_686_264_678}\\
$A B C D$ is a uniform rectangular lamina with mass $m$ and sides $A B = 6 a$ and $A D = 8 a$. The lamina rotates freely in a vertical plane about a fixed horizontal axis passing through $A$, and it is released from rest in the position with $D$ vertically above $A$. When the diagonal $A C$ makes an angle $\theta$ below the horizontal, the force acting on the lamina at $A$ has components $R$ parallel to $C A$ and $S$ perpendicular to $C A$ (see diagram).\\
(i) Find the moment of inertia of the lamina about the axis through $A$, in terms of $m$ and $a$.\\
(ii) Show that the angular speed of the lamina is $\sqrt { \frac { 3 g ( 4 + 5 \sin \theta ) } { 50 a } }$.\\
(iii) Find the angular acceleration of the lamina, in terms of $a , g$ and $\theta$.\\
(iv) Find $R$ and $S$, in terms of $m , g$ and $\theta$.
\hfill \mbox{\textit{OCR M4 2013 Q7 [14]}}