| Exam Board | OCR MEI |
|---|---|
| Module | M4 (Mechanics 4) |
| Year | 2014 |
| Session | June |
| Marks | 25 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moments of inertia |
| Type | Composite body MI calculation |
| Difficulty | Challenging +1.2 This is a multi-part Further Maths mechanics question requiring moment of inertia calculations for composite bodies, energy methods with rotating systems, and differential equations for rotational motion. While it involves several steps and FM content (making it harder than typical A-level), the techniques are standard M4 material: MI = ½mr² for cylinders, energy conservation with rotation, and setting up equations of motion. The 'show that' structure provides guidance, reducing problem-solving demand. |
| Spec | 6.03k Newton's experimental law: direct impact6.03l Newton's law: oblique impacts6.04a Centre of mass: gravitational effect |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(M_{WOOD} = \pi r^2(0.1)(700)\) and \(M_{STEEL} = \pi(0.2)^2(0.1)(7800) - \pi r^2(0.1)(7800)\) | B1 | Both masses - maybe implicit in an integral; condone working in centimetres |
| \(I_{WOOD} = 35\pi r^4\) | B1 | \(0.5(70\pi r^2)r^2\) - condone working in centimetres |
| \(I_{STEEL} = \frac{78}{125}\pi - 390\pi r^4\) | M1* | Consider difference of two \(I\)s - must be of the form \(A - Br^4\); allow derivation of \(0.5M(R^2 + r^2)\) from first principles (where M is the mass of the annulus) or \(0.5(\text{their } M_{STEEL})(r^2 + 0.2^2)\) |
| A1 | \(0.5\pi(31.2)(0.2)^2 - 0.5\pi(780r^2)r^2\) | |
| \(\frac{78}{125}\pi - 390\pi r^4 + 35\pi r^4 = 1.5\) | M1 dep* | \(I_{STEEL} + I_{WOOD} = 1.5\) |
| so \(r = 14.3\) cm to 3 sf | A1 | |
| [6] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{1}{2} \times 10 \times v^2 + \frac{1}{2} \times 20 \times v^2 + \ldots\) | B1* | KE for one (allow centimetres/metres) |
| \(\ldots 10g \times x\sin 30 - 20g \times x\) | B1* | GPE for one (allow centimetres/metres) |
| \(a = 2,\ v = 2t,\ s = t^2\) | M1 dep* | velocity and distance travelled in terms of a single variable |
| \(= \frac{1}{2}\times 10\times(2t)^2 + \frac{1}{2}\times 20\times(2t)^2 + \ldots 10g\times t^2\sin 30 - 20g\times t^2\) | ||
| \(= 87t^2\) | E1 | AG |
| \(r\omega = 2t\) with \(r = 0.25\) so \(\omega = 8t\) and \(87t^2 = \frac{1}{2}I(8t)^2\) | M1 | GPE, KE and RKE terms with conservation of energy; condone numerical slips |
| Hence \(I = \frac{87}{32}\) | E1 | |
| [6] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(36.75 - 1.875\dot{\omega} - (2\omega + k) = \frac{87}{32}\dot{\omega}\) | M1 | Use \(C = I\dot{\omega}\); correct number of terms |
| A1 | CAO | |
| \(147\int \frac{1}{64\omega + 32k - 1176}\,\mathrm{d}\omega = \int -\mathrm{d}t\) | M1 | Separate variables (not dependent on previous M mark) or using a correct integrating factor for their DE |
| \(\frac{147}{64}\ln | 64\omega + 32k - 1176 | = -t(+c)\) |
| A1 | CAO; condone not using modulus | |
| When \(t = 3,\ \omega = 24\) | B1 dep* | Use condition |
| so \(c = \frac{147}{64}\ln | 32k + 360 | + 3\) |
| \(\omega = \frac{1}{8}\left(45 + 4k\ e^{\frac{64}{147}(3-t)} + 147 - 4k\right)\) | E1 | |
| [8] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Critical value for \(45 + 4k = 0\), \(k = -11.25\) | M1, A1 | The expression \(45 + 4k\) seen in working in part (A) |
| Critical value for \(147 - 4k = 0\) | M1 | May be awarded in (B) or (C) - the expression \(147 - 4k\) seen in (B) or (C) only |
| (B) their \(-11.25 < k \leq 36.75\) | F1 | |
| (C) \(k > 36.75\) | A1 | |
| [5] |
## Question 4(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $M_{WOOD} = \pi r^2(0.1)(700)$ and $M_{STEEL} = \pi(0.2)^2(0.1)(7800) - \pi r^2(0.1)(7800)$ | B1 | Both masses - maybe implicit in an integral; condone working in centimetres |
| $I_{WOOD} = 35\pi r^4$ | B1 | $0.5(70\pi r^2)r^2$ - condone working in centimetres |
| $I_{STEEL} = \frac{78}{125}\pi - 390\pi r^4$ | M1* | Consider difference of two $I$s - must be of the form $A - Br^4$; allow derivation of $0.5M(R^2 + r^2)$ from first principles (where M is the mass of the annulus) or $0.5(\text{their } M_{STEEL})(r^2 + 0.2^2)$ |
| | A1 | $0.5\pi(31.2)(0.2)^2 - 0.5\pi(780r^2)r^2$ |
| $\frac{78}{125}\pi - 390\pi r^4 + 35\pi r^4 = 1.5$ | M1 dep* | $I_{STEEL} + I_{WOOD} = 1.5$ |
| so $r = 14.3$ cm to 3 sf | A1 | |
| **[6]** | | |
---
## Question 4(b)(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{1}{2} \times 10 \times v^2 + \frac{1}{2} \times 20 \times v^2 + \ldots$ | B1* | KE for one (allow centimetres/metres) |
| $\ldots 10g \times x\sin 30 - 20g \times x$ | B1* | GPE for one (allow centimetres/metres) |
| $a = 2,\ v = 2t,\ s = t^2$ | M1 dep* | velocity and distance travelled in terms of a single variable |
| $= \frac{1}{2}\times 10\times(2t)^2 + \frac{1}{2}\times 20\times(2t)^2 + \ldots 10g\times t^2\sin 30 - 20g\times t^2$ | | |
| $= 87t^2$ | E1 | AG |
| $r\omega = 2t$ with $r = 0.25$ so $\omega = 8t$ and $87t^2 = \frac{1}{2}I(8t)^2$ | M1 | GPE, KE and RKE terms with conservation of energy; condone numerical slips |
| Hence $I = \frac{87}{32}$ | E1 | |
| **[6]** | | |
---
## Question 4(b)(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $36.75 - 1.875\dot{\omega} - (2\omega + k) = \frac{87}{32}\dot{\omega}$ | M1 | Use $C = I\dot{\omega}$; correct number of terms |
| | A1 | CAO |
| $147\int \frac{1}{64\omega + 32k - 1176}\,\mathrm{d}\omega = \int -\mathrm{d}t$ | M1 | Separate variables (not dependent on previous M mark) or using a correct integrating factor for their DE |
| $\frac{147}{64}\ln|64\omega + 32k - 1176| = -t(+c)$ | M1* | Integrate to $A\ln(B + D\omega) = Et(+c)$ or equivalent if using IF approach |
| | A1 | CAO; condone not using modulus |
| When $t = 3,\ \omega = 24$ | B1 dep* | Use condition |
| so $c = \frac{147}{64}\ln|32k + 360| + 3$ | A1 | Correctly find $c$ |
| $\omega = \frac{1}{8}\left(45 + 4k\ e^{\frac{64}{147}(3-t)} + 147 - 4k\right)$ | E1 | |
| **[8]** | | |
---
## Question 4(b)(iii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Critical value for $45 + 4k = 0$, $k = -11.25$ | M1, A1 | The expression $45 + 4k$ seen in working in part (A) |
| Critical value for $147 - 4k = 0$ | M1 | May be awarded in (B) or (C) - the expression $147 - 4k$ seen in (B) or (C) only |
| (B) their $-11.25 < k \leq 36.75$ | F1 | |
| (C) $k > 36.75$ | A1 | |
| **[5]** | | |4
\begin{enumerate}[label=(\alph*)]
\item A pulley consists of a central cylinder of wood and an outer ring of steel. The density of the wood is $700 \mathrm {~kg} \mathrm {~m} ^ { - 3 }$ and the density of the steel is $7800 \mathrm {~kg} \mathrm {~m} ^ { - 3 }$. The pulley has a radius of 20 cm and is 10 cm thick (see Fig. 4.1).
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{c3ac9277-d34d-4d0e-9f9b-d0bce8c741af-4_359_661_404_742}
\captionsetup{labelformat=empty}
\caption{Fig. 4.1}
\end{center}
\end{figure}
Find the radius that the central cylinder must have in order that the moment of inertia of the pulley about the axis of symmetry shown in Fig. 4.1 is $1.5 \mathrm {~kg} \mathrm {~m} ^ { 2 }$.
\item Two blocks P and Q of masses 10 kg and 20 kg are connected by a light inextensible string. The string passes over a heavy rough pulley of radius 25 cm . The pulley can rotate freely and the string does not slip. Block P is held at rest in smooth contact with a plane inclined at $30 ^ { \circ }$ to the horizontal, and block Q is at rest below the pulley (see Fig. 4.2).
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{c3ac9277-d34d-4d0e-9f9b-d0bce8c741af-5_341_917_438_541}
\captionsetup{labelformat=empty}
\caption{Fig. 4.2}
\end{center}
\end{figure}
At $t \mathrm {~s}$ after the system is released from rest, the pulley has angular velocity $\omega \mathrm { rad } \mathrm { s } ^ { - 1 }$ and block P has constant acceleration of $2 \mathrm {~m} \mathrm {~s} ^ { - 2 }$ up the slope.
\begin{enumerate}[label=(\roman*)]
\item Show that the net loss of energy of the two blocks in the first $t$ seconds of motion is $87 t ^ { 2 } \mathrm {~J}$ and use the principle of conservation of energy to show that the moment of inertia of the pulley about its axis of rotation is $\frac { 87 } { 32 } \mathrm {~kg} \mathrm {~m} ^ { 2 }$.
When $t = 3$ a resistive couple is applied to the pulley. This resistive couple has magnitude $( 2 \omega + k ) \mathrm { Nm }$, where $k$ is a constant. The couple on the pulley due to tensions in the sections of string is $\left( \frac { 147 } { 4 } - \frac { 15 } { 8 } \frac { \mathrm {~d} \omega } { \mathrm {~d} t } \right) \mathrm { Nm }$ in the direction of positive $\omega$.
\item Write down a first order differential equation for $\omega$ when $t \geqslant 3$ and show by integration that
$$\omega = \frac { 1 } { 8 } \left( ( 45 + 4 k ) \mathrm { e } ^ { \frac { 64 } { 147 } ( 3 - t ) } + 147 - 4 k \right) .$$
\item By considering the equation given in part (ii), find the value or set of values of $k$ for which the pulley\\
(A) continues to rotate with constant angular velocity,\\
(B) rotates with decreasing angular velocity without coming to rest,\\
(C) rotates with decreasing angular velocity and comes to rest if there is sufficient distance between P and the pulley.
\section*{END OF QUESTION PAPER}
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{OCR MEI M4 2014 Q4 [25]}}