| Exam Board | OCR |
|---|---|
| Module | M4 (Mechanics 4) |
| Year | 2011 |
| 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 M4 rotation problem requiring moment of inertia calculation using parallel axis theorem, application of rotational dynamics with variable geometry, energy methods for angular speed, and force analysis involving both normal and frictional components. The multi-step nature, need to track changing geometry as the block rotates, and the final force calculation requiring vector resolution of reaction components make this substantially harder than average, though the techniques are all standard M4 material. |
| Spec | 6.04a Centre of mass: gravitational effect6.05a Angular velocity: definitions |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(XG = \sqrt{5}a\) | B1 | For \(I_G = \frac{1}{3}m\{a^2 + (3a)^2\}\) |
| \(I = \frac{1}{3}m\{a^2 + (3a)^2\} + m(\sqrt{5}a)^2\) | M1 | Using parallel axes rule |
| \(= \frac{25}{3}ma^2\) | A1 [3] | |
| OR: \(\frac{4}{3}(\frac{1}{6}m)\left((\frac{1}{2}a)^2 + a^2\right) + \frac{4}{3}(\frac{5}{6}m)\left((\frac{5}{2}a)^2 + a^2\right)\) | M1, A1 | Correct expression for \(I\) |
| \(I = \frac{25}{3}ma^2\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(mg(\sqrt{5}a) = I\alpha\) | M1 | Allow e.g. \(mg(2a) = I\alpha\) |
| \(\sqrt{5}mga = \frac{25}{3}ma^2\alpha\) | ||
| \(\alpha = \frac{3\sqrt{5}g}{25a}\) | A1ag [2] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{1}{2}I\omega^2 = mga\) | M1 | Equation involving KE and PE |
| \(\frac{25}{6}ma^2\omega^2 = mga\) | A1ft | |
| \(\omega = \sqrt{\frac{6g}{25a}}\) | A1 [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(H = m(XG)\omega^2\) | M1 | For using acceleration \(r\omega^2\) |
| \(= m(\sqrt{5}a)\left(\frac{6g}{25a}\right)\) | A1 | Or (\(F\) parallel to \(BA\), \(\theta\) is angle \(GXB\)): \(F - mg\sin\theta = m\left((AG)\omega^2\cos\theta - (AG)\alpha\sin\theta\right)\) |
| \(= \frac{6\sqrt{5}}{25}mg\) | A1ft | ft from incorrect \(\omega\) only; Or \(F = \frac{mg(2\sqrt{5}+12)}{25}\) |
| \(mg - V = m(XG)\alpha\) | M1 | For using acceleration \(r\alpha\) |
| \(V = mg - m(\sqrt{5}a)\left(\frac{3\sqrt{5}g}{25a}\right)\) | A1 | Or (\(R\) parallel to \(AD\)): \(mg\cos\theta - R = m\left((AG)\omega^2\sin\theta + (AG)\alpha\cos\theta\right)\) |
| \(= \frac{2}{5}mg\) | A1 | Or \(R = \frac{mg(4\sqrt{5}-6)}{25}\) |
| Force has magnitude \(\sqrt{H^2 + V^2}\) | Or \(\sqrt{F^2 + R^2}\) | |
| \(= \frac{2}{25}mg\sqrt{(3\sqrt{5})^2 + 5^2}\) | M1 | |
| \(= \frac{2\sqrt{70}}{25}mg\) | A1ag [8] |
## Question 7:
### Part (i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $XG = \sqrt{5}a$ | B1 | For $I_G = \frac{1}{3}m\{a^2 + (3a)^2\}$ |
| $I = \frac{1}{3}m\{a^2 + (3a)^2\} + m(\sqrt{5}a)^2$ | M1 | Using parallel axes rule |
| $= \frac{25}{3}ma^2$ | A1 [3] | |
| OR: $\frac{4}{3}(\frac{1}{6}m)\left((\frac{1}{2}a)^2 + a^2\right) + \frac{4}{3}(\frac{5}{6}m)\left((\frac{5}{2}a)^2 + a^2\right)$ | M1, A1 | Correct expression for $I$ |
| $I = \frac{25}{3}ma^2$ | A1 | |
### Part (ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $mg(\sqrt{5}a) = I\alpha$ | M1 | Allow e.g. $mg(2a) = I\alpha$ |
| $\sqrt{5}mga = \frac{25}{3}ma^2\alpha$ | | |
| $\alpha = \frac{3\sqrt{5}g}{25a}$ | A1ag [2] | |
### Part (iii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{1}{2}I\omega^2 = mga$ | M1 | Equation involving KE and PE |
| $\frac{25}{6}ma^2\omega^2 = mga$ | A1ft | |
| $\omega = \sqrt{\frac{6g}{25a}}$ | A1 [3] | |
### Part (iv):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $H = m(XG)\omega^2$ | M1 | For using acceleration $r\omega^2$ |
| $= m(\sqrt{5}a)\left(\frac{6g}{25a}\right)$ | A1 | Or ($F$ parallel to $BA$, $\theta$ is angle $GXB$): $F - mg\sin\theta = m\left((AG)\omega^2\cos\theta - (AG)\alpha\sin\theta\right)$ |
| $= \frac{6\sqrt{5}}{25}mg$ | A1ft | ft from incorrect $\omega$ only; Or $F = \frac{mg(2\sqrt{5}+12)}{25}$ |
| $mg - V = m(XG)\alpha$ | M1 | For using acceleration $r\alpha$ |
| $V = mg - m(\sqrt{5}a)\left(\frac{3\sqrt{5}g}{25a}\right)$ | A1 | Or ($R$ parallel to $AD$): $mg\cos\theta - R = m\left((AG)\omega^2\sin\theta + (AG)\alpha\cos\theta\right)$ |
| $= \frac{2}{5}mg$ | A1 | Or $R = \frac{mg(4\sqrt{5}-6)}{25}$ |
| Force has magnitude $\sqrt{H^2 + V^2}$ | | Or $\sqrt{F^2 + R^2}$ |
| $= \frac{2}{25}mg\sqrt{(3\sqrt{5})^2 + 5^2}$ | M1 | |
| $= \frac{2\sqrt{70}}{25}mg$ | A1ag [8] | |7\\
\includegraphics[max width=\textwidth, alt={}, center]{337dd1f9-a691-4e99-9aa7-7a93d8bb13be-3_479_1225_1484_461}
A uniform rectangular block of mass $m$ and cross-section $A B C D$ has $A B = C D = 6 a$ and $A D = B C = 2 a$. The point $X$ is on $A B$ such that $A X = a$ and $G$ is the centre of $A B C D$. The block is placed with $A B$ perpendicular to the straight edge of a rough horizontal table. $A X$ is in contact with the table and $X B$ overhangs the edge (see diagram). The block is released from rest in this position, and it rotates without slipping about a horizontal axis through $X$.\\
(i) Find the moment of inertia of the block about the axis of rotation.
For the instant when $X G$ is horizontal,\\
(ii) show that the angular acceleration of the block is $\frac { 3 \sqrt { 5 } g } { 25 a }$,\\
(iii) find the angular speed of the block,\\
(iv) show that the force exerted by the table on the block has magnitude $\frac { 2 \sqrt { 70 } } { 25 } m g$.
\hfill \mbox{\textit{OCR M4 2011 Q7 [16]}}