| Exam Board | OCR |
|---|---|
| Module | M4 (Mechanics 4) |
| Year | 2010 |
| Session | June |
| Marks | 16 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moments of inertia |
| Type | Force at pivot/axis |
| Difficulty | Challenging +1.8 This is a challenging Further Maths M4 rotation dynamics problem requiring: (i) parallel axis theorem for moment of inertia of a rectangle, (ii) energy conservation to find angular speed and torque equation for angular acceleration, (iii) resolving reaction forces in rotating frame with centripetal and tangential components. The multi-step nature, need for careful coordinate resolution, and specific numerical results (requiring precise calculation) make this significantly harder than average, though the techniques are standard for M4. |
| Spec | 6.04a Centre of mass: gravitational effect6.05a Angular velocity: definitions |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| \(I = \frac{1}{3}m\{(3a)^2 + (4a)^2\} + m(5a)^2\) | M1 | Using parallel (or perpendicular) axes rule |
| \(= \frac{100ma^2}{3}\) | A1 | or \(I = \frac{4}{3}m(3a)^2 + \frac{4}{3}m(4a)^2\) |
| A1 [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| \(\frac{1}{2}\left(\frac{100}{3}ma^2\right)\omega^2 = mg(4a - 3a)\) | M1, A1ft | Equation involving KE and PE |
| \(\frac{50}{3}ma^2\omega^2 = mga\) | ||
| Angular speed is \(\sqrt{\frac{3g}{50a}}\) | A1 | |
| \(-mg(3a) = \left(\frac{100}{3}ma^2\right)\alpha\) | ag, M1 | Using \(C = I\alpha\) |
| Angular acceleration is \((-)\dfrac{9g}{100a}\) | A1 [5] |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| \(P - mg\cos\theta = m(5a)\omega^2\) | M1 | Equation involving \(P\) and \(r\omega^2\) |
| \(P - \frac{4}{5}mg = m(5a)\left(\frac{3g}{50a}\right)\) | A2 | Give A1 if correct apart from sign(s); allow \(\frac{3}{5}H + \frac{4}{5}V\) in place of \(P\) |
| \(P = \frac{11}{10}mg\) | ||
| \(Q - mg\sin\theta = m(5a)\alpha\) | M1 | Equation involving \(Q\) and \(r\alpha\) |
| \(Q - \frac{3}{5}mg = -m(5a)\left(\frac{9g}{100a}\right)\) | A2ft | Give A1 if correct apart from sign(s); ft for wrong value of \(\alpha\) or wrong value of \(r\); allow \(\frac{3}{5}V - \frac{4}{5}H\) in place of \(Q\) |
| \(Q = \frac{3}{20}mg\) | ||
| \(F = \sqrt{P^2 + Q^2} = \frac{1}{20}mg\sqrt{22^2 + 3^2}\) | M1 | Dependent on previous M1M1 |
| \(= \dfrac{\sqrt{493}}{20}mg\) | A1, ag [8] |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| \(H = m(5a)\omega^2\sin\theta - m(5a)\alpha\cos\theta\) | M1 | Equation involving \(H\), \(r\omega^2\) and \(r\alpha\) |
| \(H = m(5a)\left(\frac{3g}{50a}\right)\left(\frac{3}{5}\right) + m(5a)\left(\frac{9g}{100a}\right)\left(\frac{4}{5}\right)\) | A2ft | Give A1 if correct apart from sign(s) |
| \(V - mg = m(5a)\omega^2\cos\theta + m(5a)\alpha\sin\theta\) | M1 | Equation involving \(V\), \(r\omega^2\) and \(r\alpha\) |
| \(V - mg = m(5a)\left(\frac{3g}{50a}\right)\left(\frac{4}{5}\right) - m(5a)\left(\frac{9g}{100a}\right)\left(\frac{3}{5}\right)\) | A2ft | Give A1 if correct apart from sign(s) |
| \(H = \frac{27}{50}mg\), \(V = \frac{97}{100}mg\) | ||
| \(F = \sqrt{H^2+V^2} = \frac{1}{100}mg\sqrt{54^2+97^2} = \frac{\sqrt{12325}}{100}mg = \frac{\sqrt{493}}{20}mg\) | M1, A1, ag | Dependent on previous M1M1 |
# Question 7:
## Part (i):
| Working/Answer | Marks | Guidance |
|---|---|---|
| $I = \frac{1}{3}m\{(3a)^2 + (4a)^2\} + m(5a)^2$ | M1 | Using parallel (or perpendicular) axes rule |
| $= \frac{100ma^2}{3}$ | A1 | or $I = \frac{4}{3}m(3a)^2 + \frac{4}{3}m(4a)^2$ |
| | A1 **[3]** | |
## Part (ii):
| Working/Answer | Marks | Guidance |
|---|---|---|
| $\frac{1}{2}\left(\frac{100}{3}ma^2\right)\omega^2 = mg(4a - 3a)$ | M1, A1ft | Equation involving KE and PE |
| $\frac{50}{3}ma^2\omega^2 = mga$ | | |
| Angular speed is $\sqrt{\frac{3g}{50a}}$ | A1 | |
| $-mg(3a) = \left(\frac{100}{3}ma^2\right)\alpha$ | ag, M1 | Using $C = I\alpha$ |
| Angular acceleration is $(-)\dfrac{9g}{100a}$ | A1 **[5]** | |
## Part (iii):
| Working/Answer | Marks | Guidance |
|---|---|---|
| $P - mg\cos\theta = m(5a)\omega^2$ | M1 | Equation involving $P$ and $r\omega^2$ |
| $P - \frac{4}{5}mg = m(5a)\left(\frac{3g}{50a}\right)$ | A2 | Give A1 if correct apart from sign(s); allow $\frac{3}{5}H + \frac{4}{5}V$ in place of $P$ |
| $P = \frac{11}{10}mg$ | | |
| $Q - mg\sin\theta = m(5a)\alpha$ | M1 | Equation involving $Q$ and $r\alpha$ |
| $Q - \frac{3}{5}mg = -m(5a)\left(\frac{9g}{100a}\right)$ | A2ft | Give A1 if correct apart from sign(s); ft for wrong value of $\alpha$ or wrong value of $r$; allow $\frac{3}{5}V - \frac{4}{5}H$ in place of $Q$ |
| $Q = \frac{3}{20}mg$ | | |
| $F = \sqrt{P^2 + Q^2} = \frac{1}{20}mg\sqrt{22^2 + 3^2}$ | M1 | Dependent on previous M1M1 |
| $= \dfrac{\sqrt{493}}{20}mg$ | A1, ag **[8]** | |
### Alternative for (iii):
| Working/Answer | Marks | Guidance |
|---|---|---|
| $H = m(5a)\omega^2\sin\theta - m(5a)\alpha\cos\theta$ | M1 | Equation involving $H$, $r\omega^2$ and $r\alpha$ |
| $H = m(5a)\left(\frac{3g}{50a}\right)\left(\frac{3}{5}\right) + m(5a)\left(\frac{9g}{100a}\right)\left(\frac{4}{5}\right)$ | A2ft | Give A1 if correct apart from sign(s) |
| $V - mg = m(5a)\omega^2\cos\theta + m(5a)\alpha\sin\theta$ | M1 | Equation involving $V$, $r\omega^2$ and $r\alpha$ |
| $V - mg = m(5a)\left(\frac{3g}{50a}\right)\left(\frac{4}{5}\right) - m(5a)\left(\frac{9g}{100a}\right)\left(\frac{3}{5}\right)$ | A2ft | Give A1 if correct apart from sign(s) |
| $H = \frac{27}{50}mg$, $V = \frac{97}{100}mg$ | | |
| $F = \sqrt{H^2+V^2} = \frac{1}{100}mg\sqrt{54^2+97^2} = \frac{\sqrt{12325}}{100}mg = \frac{\sqrt{493}}{20}mg$ | M1, A1, ag | Dependent on previous M1M1 |7\\
\includegraphics[max width=\textwidth, alt={}, center]{ea62d6d9-ac2f-44e7-8bfb-ae9aeea7109b-4_524_732_258_705}
The diagram shows a uniform rectangular lamina $A B C D$ with $A B = 6 a , A D = 8 a$ and centre $G$. The mass of the lamina is $m$. The lamina rotates freely in a vertical plane about a fixed horizontal axis passing through $A$ and perpendicular to the lamina.\\
(i) Find the moment of inertia of the lamina about this axis.
The lamina is released from rest with $A D$ horizontal and $B C$ below $A D$.\\
(ii) For an instant during the subsequent motion when $A D$ is vertical, show that the angular speed of the lamina is $\sqrt { \frac { 3 g } { 50 a } }$ and find its angular acceleration.
At an instant when $A D$ is vertical, the force acting on the lamina at $A$ has magnitude $F$.\\
(iii) By finding components parallel and perpendicular to $G A$, or otherwise, show that $F = \frac { \sqrt { 493 } } { 20 } \mathrm { mg }$.\\[0pt]
[8]
\hfill \mbox{\textit{OCR M4 2010 Q7 [16]}}