| Exam Board | OCR |
|---|---|
| Module | M4 (Mechanics 4) |
| Year | 2015 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moments of inertia |
| Type | Prove MI by integration |
| Difficulty | Challenging +1.8 This is a multi-part M4 question requiring integration to derive a 3D moment of inertia formula using the perpendicular axis theorem, then applying the parallel axis theorem. The derivation involves setting up an integral over disc elements, careful algebraic manipulation, and understanding of advanced mechanics concepts. While systematic, it requires strong technical facility with integration and MOI theorems beyond standard A-level, placing it well above average difficulty. |
| Spec | 1.08a Fundamental theorem of calculus: integration as reverse of differentiation1.08b Integrate x^n: where n != -1 and sums6.04e Rigid body equilibrium: coplanar forces |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{1}{4}m(2a)^2\ (= ma^2)\) | B1 [1] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Mass per unit volume is \(\rho = \dfrac{M}{\pi(2r)^2 h}\) | B1 | |
| MI of elemental disc about a diameter is \((4\pi r^2 \delta x \rho)r^2\) | B1 | \((Mr^2/h)\delta x\) (condone lack of \(\delta x\)) |
| MI of elemental disc about an end face is \(4\rho\pi r^4\delta x + (4\rho\pi r^2\delta x)x^2\) | *M1 A1 | \(\frac{M}{h}(r^2 + x^2)\delta x\) - M1 for using parallel axes rule – must be of the form \(\lambda r^4 + \mu r^2 x^2\) (condone lack of \(\delta x\)) |
| \(I = 4\rho\pi r^2\int_0^h (r^2 + x^2)\,dx\) | A1 | \(\frac{M}{h}\int_0^h r^2 + x^2\,dx\) |
| \(= 4\rho\pi r^2\left[r^2x + \frac{1}{3}x^3\right]_0^h = 4\left(\frac{M}{4\pi r^2 h}\right)\pi r^4 h + \frac{4}{3}\left(\frac{M}{4\pi r^2 h}\right)\pi r^2 h^3\) | M1 dep* | Integrating and obtaining an expression for \(I\) in terms \(M\), \(r\) and \(h\) – must be using the correct limits |
| \(= M\left(r^2 + \frac{1}{3}h^2\right)\) | A1 | AG www correctly obtained – if \(\delta x\) is omitted throughout then do not award final A mark |
| MI through the centre of the cylinder \(= M\left(r^2 + \frac{1}{3}h^2\right) - M\left(\frac{h}{2}\right)^2\ \left(= M\left(r^2 + \frac{1}{12}h^2\right)\right)\) | B1 [8] |
# Question 4:
## Part (i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{1}{4}m(2a)^2\ (= ma^2)$ | B1 **[1]** | |
## Part (ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Mass per unit volume is $\rho = \dfrac{M}{\pi(2r)^2 h}$ | B1 | |
| MI of elemental disc about a diameter is $(4\pi r^2 \delta x \rho)r^2$ | B1 | $(Mr^2/h)\delta x$ (condone lack of $\delta x$) |
| MI of elemental disc about an end face is $4\rho\pi r^4\delta x + (4\rho\pi r^2\delta x)x^2$ | *M1 A1 | $\frac{M}{h}(r^2 + x^2)\delta x$ - M1 for using parallel axes rule – must be of the form $\lambda r^4 + \mu r^2 x^2$ (condone lack of $\delta x$) |
| $I = 4\rho\pi r^2\int_0^h (r^2 + x^2)\,dx$ | A1 | $\frac{M}{h}\int_0^h r^2 + x^2\,dx$ |
| $= 4\rho\pi r^2\left[r^2x + \frac{1}{3}x^3\right]_0^h = 4\left(\frac{M}{4\pi r^2 h}\right)\pi r^4 h + \frac{4}{3}\left(\frac{M}{4\pi r^2 h}\right)\pi r^2 h^3$ | M1 dep* | Integrating and obtaining an expression for $I$ in terms $M$, $r$ and $h$ – must be using the correct limits |
| $= M\left(r^2 + \frac{1}{3}h^2\right)$ | A1 | **AG www** correctly obtained – if $\delta x$ is omitted throughout then do not award final A mark |
| MI through the centre of the cylinder $= M\left(r^2 + \frac{1}{3}h^2\right) - M\left(\frac{h}{2}\right)^2\ \left(= M\left(r^2 + \frac{1}{12}h^2\right)\right)$ | B1 **[8]** | |
---4 (i) Write down the moment of inertia of a uniform circular disc of mass $m$ and radius $2 a$ about a diameter.
A uniform solid cylinder has mass $M$, radius $2 r$ and height $h$.\\
(ii) Show by integration, and using the result from part (i), that the moment of inertia of the cylinder about a diameter of an end face is
$$M \left( r ^ { 2 } + \frac { 1 } { 3 } h ^ { 2 } \right)$$
and hence find the moment of inertia of the cylinder about a diameter through the centre of the cylinder.\\
\includegraphics[max width=\textwidth, alt={}, center]{4b50b084-081f-48d2-ad5b-95b2c9e55dfc-3_919_897_260_591}
A smooth circular wire hoop, with centre $O$ and radius $r$, is fixed in a vertical plane. The highest point on the wire is $H$. A small bead $B$ of mass $m$ is free to move along the wire. A light inextensible string of length $a$, where $a > 2 r$, has one end attached to the bead. The other end of the string passes over a small smooth pulley at $H$ and carries at its end a particle $P$ of mass $\lambda m$, where $\lambda$ is a positive constant. The part of the string $H P$ is vertical and the part of the string $B H$ makes an angle $\theta$ radians with the downward vertical where $0 \leqslant \theta \leqslant \frac { 1 } { 3 } \pi$ (see diagram). You may assume that $P$ remains above the lowest point of the wire.\\
\hfill \mbox{\textit{OCR M4 2015 Q4 [9]}}