| Exam Board | OCR |
|---|---|
| Module | M4 (Mechanics 4) |
| Year | 2004 |
| Session | June |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moments of inertia |
| Type | Prove MI by integration |
| Difficulty | Standard +0.8 This is a multi-part mechanics question requiring integration to derive moment of inertia, then applying rotational dynamics with forces at two different positions. While the integration is straightforward, parts (ii) and (iii) require careful application of Newton's second law in rotational form, resolving forces, and understanding the distinction between initial angular acceleration and later motion. The non-symmetric pivot point adds complexity beyond standard textbook examples. |
| Spec | 6.04d Integration: for centre of mass of laminas/solids6.04e Rigid body equilibrium: coplanar forces6.05f Vertical circle: motion including free fall |
| Answer | Marks | Guidance |
|---|---|---|
| \(I = \sum(\rho\,\delta x)x^2 = \int_{-\frac{2}{3}a}^{\frac{4}{3}a} \frac{m}{2a}x^2\,dx\) | M1, A1 | |
| \(= \left[\frac{mx^3}{6a}\right]_{-\frac{2}{3}a}^{\frac{4}{3}a} = \frac{32}{81}ma^2 + \frac{4}{81}ma^2\) | M1 | Dependent on previous M1 |
| \(= \frac{4}{9}ma^2\) | A1 (ag) | Total: 4 |
| OR \(I_G = \sum(\rho\,\delta x)x^2 = \int_{-a}^{a}\frac{m}{2a}x^2\,dx\) | M1A1 | Or \(2\int_0^a \frac{m}{2a}x^2\,dx\) |
| \(I = \frac{1}{6}ma^2 + \frac{1}{6}ma^2 + m(\frac{1}{3}a)^2\) | M1 | Dependent on previous M1 |
| \(= \frac{4}{9}ma^2\) | A1 (ag) |
| Answer | Marks | Guidance |
|---|---|---|
| \(mg(\frac{1}{3}a) = (\frac{4}{9}ma^2)\alpha\) | M1 | Use of \(L = I\alpha\) |
| \(\alpha = \frac{3g}{4a}\) | A1 | |
| \(mg - R = m(\frac{1}{3}a)\alpha\) | M1 | For force \(= m(\frac{1}{3}a)\alpha\) |
| \(mg - R = \frac{1}{4}mg\) | A1 ft | |
| \(R = \frac{3}{4}mg\) vertically upwards | A1 | Direction must be indicated Total: 5 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{1}{2}(\frac{4}{9}ma^2)\omega^2 = mg(\frac{1}{3}a)\) | M1 | Use of \(\frac{1}{2}I\omega^2\) |
| \(\omega^2 = \frac{3g}{2a}\) | A1 | |
| \(S - mg = m(\frac{1}{3}a)\omega^2\) | M1 | For force \(= m(\frac{1}{3}a)\omega^2\) |
| \(S - mg = \frac{1}{2}mg\) | A1 ft | |
| \(S = \frac{3}{2}mg\) vertically upwards | A1 | Direction must be indicated Total: 5 |
| Answer | Marks |
|---|---|
| \(\ddot{\theta} = \frac{3g\cos\theta}{4a}\) | M1A1 |
| \(\dot{\theta}^2 = \frac{3g\sin\theta}{2a}\) | M1A1 |
| \(mg\cos\theta - Y = m(\frac{1}{3}a)\frac{3g\cos\theta}{4a}\), \(Y = \frac{3}{4}mg\cos\theta\) | M1A1 ft |
| \(X - mg\sin\theta = m(\frac{1}{3}a)\frac{3g\sin\theta}{2a}\), \(X = \frac{3}{2}mg\sin\theta\) | M1A1 ft |
| When \(\theta = 0\): \(X = 0\), \(Y = \frac{3}{4}mg\) | A1 |
| When \(\theta = \frac{1}{2}\pi\): \(X = \frac{3}{2}mg\), \(Y = 0\) | A1 |
# Question 7(i):
$I = \sum(\rho\,\delta x)x^2 = \int_{-\frac{2}{3}a}^{\frac{4}{3}a} \frac{m}{2a}x^2\,dx$ | M1, A1 |
$= \left[\frac{mx^3}{6a}\right]_{-\frac{2}{3}a}^{\frac{4}{3}a} = \frac{32}{81}ma^2 + \frac{4}{81}ma^2$ | M1 | Dependent on previous M1
$= \frac{4}{9}ma^2$ | A1 (ag) | **Total: 4**
OR $I_G = \sum(\rho\,\delta x)x^2 = \int_{-a}^{a}\frac{m}{2a}x^2\,dx$ | M1A1 | Or $2\int_0^a \frac{m}{2a}x^2\,dx$
$I = \frac{1}{6}ma^2 + \frac{1}{6}ma^2 + m(\frac{1}{3}a)^2$ | M1 | Dependent on previous M1
$= \frac{4}{9}ma^2$ | A1 (ag) |
## Question 7(ii):
$mg(\frac{1}{3}a) = (\frac{4}{9}ma^2)\alpha$ | M1 | Use of $L = I\alpha$
$\alpha = \frac{3g}{4a}$ | A1 |
$mg - R = m(\frac{1}{3}a)\alpha$ | M1 | For force $= m(\frac{1}{3}a)\alpha$
$mg - R = \frac{1}{4}mg$ | A1 ft |
$R = \frac{3}{4}mg$ vertically upwards | A1 | Direction must be indicated **Total: 5**
## Question 7(iii):
$\frac{1}{2}(\frac{4}{9}ma^2)\omega^2 = mg(\frac{1}{3}a)$ | M1 | Use of $\frac{1}{2}I\omega^2$
$\omega^2 = \frac{3g}{2a}$ | A1 |
$S - mg = m(\frac{1}{3}a)\omega^2$ | M1 | For force $= m(\frac{1}{3}a)\omega^2$
$S - mg = \frac{1}{2}mg$ | A1 ft |
$S = \frac{3}{2}mg$ vertically upwards | A1 | Direction must be indicated **Total: 5**
Alternative for (ii) and (iii): ($\theta$ measured from horizontal; $X$ is radial component; $Y$ is transverse component)
$\ddot{\theta} = \frac{3g\cos\theta}{4a}$ | M1A1 |
$\dot{\theta}^2 = \frac{3g\sin\theta}{2a}$ | M1A1 |
$mg\cos\theta - Y = m(\frac{1}{3}a)\frac{3g\cos\theta}{4a}$, $Y = \frac{3}{4}mg\cos\theta$ | M1A1 ft |
$X - mg\sin\theta = m(\frac{1}{3}a)\frac{3g\sin\theta}{2a}$, $X = \frac{3}{2}mg\sin\theta$ | M1A1 ft |
When $\theta = 0$: $X = 0$, $Y = \frac{3}{4}mg$ | A1 |
When $\theta = \frac{1}{2}\pi$: $X = \frac{3}{2}mg$, $Y = 0$ | A1 |7 A uniform rod $A B$ has mass $m$ and length $2 a$. The point $P$ on the rod is such that $A P = \frac { 2 } { 3 } a$.\\
(i) Prove by integration that the moment of inertia of the rod about an axis through $P$ perpendicular to $A B$ is $\frac { 4 } { 9 } m a ^ { 2 }$.
The axis through $P$ is fixed and horizontal, and the rod can rotate without resistance in a vertical plane about this axis. The rod is released from rest in a horizontal position. Find, in terms of $m$ and $g$,\\
(ii) the force acting on the rod at $P$ immediately after the release of the rod,\\
(iii) the force acting on the rod at $P$ at an instant in the subsequent motion when $B$ is vertically below $P$.
\hfill \mbox{\textit{OCR M4 2004 Q7 [14]}}