| Exam Board | OCR |
|---|---|
| Module | M4 (Mechanics 4) |
| Year | 2014 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moments of inertia |
| Type | Prove MI by integration |
| Difficulty | Challenging +1.2 This is a standard M4 mechanics question requiring integration to derive moment of inertia (with formula provided), then applying rotational dynamics equations. The integration setup is straightforward using disc elements, and parts (ii)-(iii) are routine applications of τ = Iα and kinematic equations. More challenging than basic mechanics due to the integration and rotational concepts, but follows a well-established template for this topic. |
| Spec | 1.08a Fundamental theorem of calculus: integration as reverse of differentiation1.08b Integrate x^n: where n != -1 and sums3.04a Calculate moments: about a point3.04b Equilibrium: zero resultant moment and force6.04e 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 |
| Mass per unit volume \(\rho = \frac{M}{\frac{1}{3}\pi R^2 h}\) | B1 | |
| \(I = \sum \frac{1}{2}(\rho\pi y^2 \delta x)y^2 = \frac{1}{2}\rho\pi\int y^4\,dx\) | M1 | M1 for \(\frac{1}{2}\int y^4\,dx\) |
| \(= \frac{1}{2}\rho\pi\int_0^h \frac{R^4x^4}{h^4}\,dx\) | M1A1 | M1 for substituting \(y = \frac{R}{h}x\); A1 for correct integral with limits |
| \(= \frac{1}{2}\rho\pi\left[\frac{R^4x^5}{5h^4}\right]_0^h\) | A1 | A1 for correct integration |
| \(= \frac{1}{10}\left(\frac{3M}{\pi R^2 h}\right)\pi R^4 h = \frac{3}{10}MR^2\) | A1 | AG Correctly shown |
| [6] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| MI of cone \(= \frac{3}{10}(2)(0.3)^2\) | B1 | Use of \(\frac{3}{10}mr^2\) with \(m=2\) and \(r=0.3\) |
| \(0.027 = 0.054\alpha\) | M1 | Using \(C = I\alpha\) |
| \(\alpha = 0.5\) rad \(s^{-2}\) | A1 | |
| [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(6 - 0.5t = 0\) | M1 | Use of \(\omega = \omega_0 + \alpha t\) |
| \(t = 12\,s\) | A1 | cao |
| [2] |
# Question 2:
## Part (i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Mass per unit volume $\rho = \frac{M}{\frac{1}{3}\pi R^2 h}$ | B1 | |
| $I = \sum \frac{1}{2}(\rho\pi y^2 \delta x)y^2 = \frac{1}{2}\rho\pi\int y^4\,dx$ | M1 | M1 for $\frac{1}{2}\int y^4\,dx$ |
| $= \frac{1}{2}\rho\pi\int_0^h \frac{R^4x^4}{h^4}\,dx$ | M1A1 | M1 for substituting $y = \frac{R}{h}x$; A1 for correct integral with limits |
| $= \frac{1}{2}\rho\pi\left[\frac{R^4x^5}{5h^4}\right]_0^h$ | A1 | A1 for correct integration |
| $= \frac{1}{10}\left(\frac{3M}{\pi R^2 h}\right)\pi R^4 h = \frac{3}{10}MR^2$ | A1 | **AG** Correctly shown |
| **[6]** | | |
## Part (ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| MI of cone $= \frac{3}{10}(2)(0.3)^2$ | B1 | Use of $\frac{3}{10}mr^2$ with $m=2$ and $r=0.3$ |
| $0.027 = 0.054\alpha$ | M1 | Using $C = I\alpha$ |
| $\alpha = 0.5$ rad $s^{-2}$ | A1 | |
| **[3]** | | |
## Part (iii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $6 - 0.5t = 0$ | M1 | Use of $\omega = \omega_0 + \alpha t$ |
| $t = 12\,s$ | A1 | cao |
| **[2]** | | |
---2 A uniform solid circular cone has mass $M$ and base radius $R$.\\
(i) Show by integration that the moment of inertia of the cone about its axis of symmetry is $\frac { 3 } { 10 } M R ^ { 2 }$. (You may assume the standard formula $\frac { 1 } { 2 } m r ^ { 2 }$ for the moment of inertia of a uniform disc about its axis and that the volume of a cone is $\frac { 1 } { 3 } \pi r ^ { 2 } h$.)
The axis of symmetry of the cone is fixed vertically and the cone is rotating about its axis at an angular speed of $6 \mathrm { rad } \mathrm { s } ^ { - 1 }$. A frictional couple of constant moment 0.027 Nm is applied to the cone bringing it to rest. Given that the mass of the cone is 2 kg and its base radius is 0.3 m , find\\
(ii) the constant angular deceleration of the cone,\\
(iii) the time taken for the cone to come to rest from the instant that the couple is applied.
\hfill \mbox{\textit{OCR M4 2014 Q2 [11]}}