| Exam Board | OCR |
|---|---|
| Module | M4 (Mechanics 4) |
| Year | 2011 |
| 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 compound pendulum question requiring integration to find moment of inertia (with the disc formula given), then applying the compound pendulum formula T = 2π√(I/mgh). The integration is straightforward using discs, and the small oscillations formula is bookwork. More challenging than basic C1/C2 but routine for M4 students who know the standard techniques. |
| Spec | 6.04a Centre of mass: gravitational effect6.05a Angular velocity: definitions |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \((\frac{4}{3}\pi a^3)\rho = 10M\), so \(\rho = \frac{15M}{2\pi a^3}\) | M1 | |
| \(I = \sum \frac{1}{2}(\rho\pi y^2 \delta x)y^2 = \frac{1}{2}\rho\pi \int y^4 \mathrm{d}x\) | M1 | For \(\int y^4 \mathrm{d}x\) |
| \(= \frac{1}{2}\rho\pi \int_{-a}^{a}(a^2-x^2)^2 \mathrm{d}x\) | A1 | Correct integral expression including limits |
| \(= \frac{1}{2}\rho\pi\left[a^4x - \frac{2}{3}a^2x^3 + \frac{1}{5}x^5\right]_{-a}^{a}\) | A1 | For \(a^4x - \frac{2}{3}a^2x^3 + \frac{1}{5}x^5\) |
| \(= \frac{1}{2}\rho\pi\left(a^5 - \frac{2}{3}a^5 + \frac{1}{5}a^5\right) \times 2\) | ||
| \(= \frac{8}{15}\rho\pi a^5\) | A1 | |
| \(= \frac{8}{15} \times \frac{15M}{2\pi a^3} \times \pi a^5 = 4Ma^2\) | A1 ag | |
| [6] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| MI is \(4Ma^2 + Ma^2 = 5Ma^2\) | M1, A1 | |
| \(-Mga\sin\theta = 5Ma^2\ddot{\theta}\) | M1 | Equation of motion |
| \(\ddot{\theta} \approx -\frac{g}{5a}\theta\) | M1 | Obtaining period |
| Period is \(2\pi\sqrt{\frac{5a}{g}}\) | A1 | |
| [5] | ||
| *Alternative for last 3 marks:* | ||
| \(11M\bar{x} = 10M(0) + Ma\), \(\bar{x} = \frac{1}{11}a\) | M1 | Finding centre of mass |
| Period is \(2\pi\sqrt{\frac{I}{mgh}} = 2\pi\sqrt{\frac{5Ma^2}{11Mg\frac{1}{11}a}} = 2\pi\sqrt{\frac{5a}{g}}\) | M1, A1 | Using formula. *Note \(2\pi\sqrt{\frac{I}{Mgh}} = 2\pi\sqrt{\frac{5Ma^2}{Mga}}\) is M0* |
# Question 5:
## Part (i)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $(\frac{4}{3}\pi a^3)\rho = 10M$, so $\rho = \frac{15M}{2\pi a^3}$ | M1 | |
| $I = \sum \frac{1}{2}(\rho\pi y^2 \delta x)y^2 = \frac{1}{2}\rho\pi \int y^4 \mathrm{d}x$ | M1 | For $\int y^4 \mathrm{d}x$ |
| $= \frac{1}{2}\rho\pi \int_{-a}^{a}(a^2-x^2)^2 \mathrm{d}x$ | A1 | Correct integral expression including limits |
| $= \frac{1}{2}\rho\pi\left[a^4x - \frac{2}{3}a^2x^3 + \frac{1}{5}x^5\right]_{-a}^{a}$ | A1 | For $a^4x - \frac{2}{3}a^2x^3 + \frac{1}{5}x^5$ |
| $= \frac{1}{2}\rho\pi\left(a^5 - \frac{2}{3}a^5 + \frac{1}{5}a^5\right) \times 2$ | | |
| $= \frac{8}{15}\rho\pi a^5$ | A1 | |
| $= \frac{8}{15} \times \frac{15M}{2\pi a^3} \times \pi a^5 = 4Ma^2$ | A1 ag | |
| | [6] | |
## Part (ii)
| Answer/Working | Marks | Guidance |
|---|---|---|
| MI is $4Ma^2 + Ma^2 = 5Ma^2$ | M1, A1 | |
| $-Mga\sin\theta = 5Ma^2\ddot{\theta}$ | M1 | Equation of motion |
| $\ddot{\theta} \approx -\frac{g}{5a}\theta$ | M1 | Obtaining period |
| Period is $2\pi\sqrt{\frac{5a}{g}}$ | A1 | |
| | [5] | |
| *Alternative for last 3 marks:* | | |
| $11M\bar{x} = 10M(0) + Ma$, $\bar{x} = \frac{1}{11}a$ | M1 | Finding centre of mass |
| Period is $2\pi\sqrt{\frac{I}{mgh}} = 2\pi\sqrt{\frac{5Ma^2}{11Mg\frac{1}{11}a}} = 2\pi\sqrt{\frac{5a}{g}}$ | M1, A1 | Using formula. *Note $2\pi\sqrt{\frac{I}{Mgh}} = 2\pi\sqrt{\frac{5Ma^2}{Mga}}$ is M0* |
---5 The region inside the circle $x ^ { 2 } + y ^ { 2 } = a ^ { 2 }$ is rotated about the $x$-axis to form a uniform solid sphere of radius $a$ and volume $\frac { 4 } { 3 } \pi a ^ { 3 }$. The mass of the sphere is $10 M$.\\
(i) Show by integration that the moment of inertia of the sphere about the $x$-axis is $4 M a ^ { 2 }$. (You may assume the standard formula $\frac { 1 } { 2 } m r ^ { 2 }$ for the moment of inertia of a uniform disc about its axis.)
The sphere is free to rotate about a fixed horizontal axis which is a diameter of the sphere. A particle of mass $M$ is attached to the lowest point of the sphere. The sphere with the particle attached then makes small oscillations as a compound pendulum.\\
(ii) Find, in terms of $a$ and $g$, the approximate period of these oscillations.
\hfill \mbox{\textit{OCR M4 2011 Q5 [11]}}