| Exam Board | OCR |
|---|---|
| Module | M4 (Mechanics 4) |
| Year | 2017 |
| Session | June |
| Marks | 17 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moments of inertia |
| Type | Composite body MI calculation |
| Difficulty | Challenging +1.2 This is a standard M4 potential energy and equilibrium question requiring systematic application of techniques: expressing elastic and gravitational PE, differentiating, finding equilibrium positions by setting dV/dθ=0, and using the second derivative test for stability. While it involves multiple steps and careful algebra with the elastic string geometry, it follows a well-established method taught in M4 with no novel insights required. The given answer for dV/dθ guides the working, making it more routine than exploratory. |
| 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 force6.02d Mechanical energy: KE and PE concepts6.02e Calculate KE and PE: using formulae6.02h Elastic PE: 1/2 k x^26.02i Conservation of energy: mechanical energy principle6.04e Rigid body equilibrium: coplanar forces |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| GPE for rod: \(-2mga\sin\theta\) | B1 | |
| EPE for string: \(\frac{\lambda mg(2a + 4a\sin\theta)^2}{2a}\) | M1 A1 | Genuine attempt at extension and substitution into \(\frac{\lambda x^2}{2a}\) |
| \(V = 2\lambda mga(1+2\sin\theta)^2 - 2mga\sin\theta\) | A1 | Accept unsimplified |
| \(\frac{dV}{d\theta} = 4\lambda mga(1+2\sin\theta)(2\cos\theta) - 2mga\cos\theta\) | M1 | Differentiates their \(V\) |
| \(\frac{dV}{d\theta} = 2mga\cos\theta\big(4\lambda(1+2\sin\theta)-1\big)\) | A1 [6] | AG www |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\cos\theta = 0 \Rightarrow \theta = \frac{\pi}{2}\) and \(\sin\theta = \frac{1-4\lambda}{8\lambda}\left(\Rightarrow \theta = \sin^{-1}\left(\frac{1-4\lambda}{8\lambda}\right)\right)\) | M1, A1 A1 | Set \(V' = 0\); A1 for \(\theta = \frac{\pi}{2}\), A1 for existence of root at \(\sin^{-1}\left(\frac{1-4\lambda}{8\lambda}\right)\) for \(\lambda > \frac{1}{12}\) |
| \(\frac{d^2V}{d\theta^2} = -2mga\sin\theta\big(4\lambda(1+2\sin\theta)-1\big) + \ldots + 2mga\cos\theta\big(4\lambda(2\cos\theta)\big)\) | M1 A1 | M1 Attempt to differentiate \(V'\); At least two terms correct for M mark |
| \(V'' = -2mga(12\lambda - 1)\) | M1 | Substitute their \(\theta = \frac{\pi}{2}\) into their \(V''\) |
| \(\lambda > \frac{1}{12} \Rightarrow (12\lambda-1) > 0 \therefore V'' < 0 \Rightarrow\) unstable | A1 | |
| \(V'' = 16mga\lambda\left(1 - \left(\frac{1-4\lambda}{8\lambda}\right)^2\right)\) or \(mga\left(-\frac{1}{4\lambda}+2+12\lambda\right)\) | M1 | Substitute their \(\sin\theta = \frac{1-4\lambda}{8\lambda}\) into their \(V''\) or \(V'' = 16\lambda mga\cos^2\theta\) (positive for all values of \(\theta\)) |
| \(\lambda > \frac{1}{12} \Rightarrow \left(\frac{1-4\lambda}{8\lambda}\right)^2 < 1 \therefore V'' > 0 \Rightarrow\) stable | A1 [9] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(V'' = -2mga(12\lambda - 1)\) | M1 | Substitute \(\theta = \frac{\pi}{2}\) (not their \(\theta\)) into their \(V''\) |
| \(\lambda < \frac{1}{12} \Rightarrow (12\lambda-1) < 0 \therefore V'' > 0 \Rightarrow\) stable | A1 [2] |
# Question 3:
## Part (i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| GPE for rod: $-2mga\sin\theta$ | B1 | |
| EPE for string: $\frac{\lambda mg(2a + 4a\sin\theta)^2}{2a}$ | M1 A1 | Genuine attempt at extension and substitution into $\frac{\lambda x^2}{2a}$ |
| $V = 2\lambda mga(1+2\sin\theta)^2 - 2mga\sin\theta$ | A1 | Accept unsimplified |
| $\frac{dV}{d\theta} = 4\lambda mga(1+2\sin\theta)(2\cos\theta) - 2mga\cos\theta$ | M1 | Differentiates their $V$ |
| $\frac{dV}{d\theta} = 2mga\cos\theta\big(4\lambda(1+2\sin\theta)-1\big)$ | A1 **[6]** | AG www |
## Part (ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\cos\theta = 0 \Rightarrow \theta = \frac{\pi}{2}$ and $\sin\theta = \frac{1-4\lambda}{8\lambda}\left(\Rightarrow \theta = \sin^{-1}\left(\frac{1-4\lambda}{8\lambda}\right)\right)$ | M1, A1 A1 | Set $V' = 0$; A1 for $\theta = \frac{\pi}{2}$, A1 for existence of root at $\sin^{-1}\left(\frac{1-4\lambda}{8\lambda}\right)$ for $\lambda > \frac{1}{12}$ |
| $\frac{d^2V}{d\theta^2} = -2mga\sin\theta\big(4\lambda(1+2\sin\theta)-1\big) + \ldots + 2mga\cos\theta\big(4\lambda(2\cos\theta)\big)$ | M1 A1 | M1 Attempt to differentiate $V'$; At least two terms correct for M mark |
| $V'' = -2mga(12\lambda - 1)$ | M1 | Substitute their $\theta = \frac{\pi}{2}$ into their $V''$ |
| $\lambda > \frac{1}{12} \Rightarrow (12\lambda-1) > 0 \therefore V'' < 0 \Rightarrow$ unstable | A1 | |
| $V'' = 16mga\lambda\left(1 - \left(\frac{1-4\lambda}{8\lambda}\right)^2\right)$ or $mga\left(-\frac{1}{4\lambda}+2+12\lambda\right)$ | M1 | Substitute their $\sin\theta = \frac{1-4\lambda}{8\lambda}$ into their $V''$ or $V'' = 16\lambda mga\cos^2\theta$ (positive for all values of $\theta$) |
| $\lambda > \frac{1}{12} \Rightarrow \left(\frac{1-4\lambda}{8\lambda}\right)^2 < 1 \therefore V'' > 0 \Rightarrow$ stable | A1 **[9]** | |
## Part (iii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $V'' = -2mga(12\lambda - 1)$ | M1 | Substitute $\theta = \frac{\pi}{2}$ (not their $\theta$) into their $V''$ |
| $\lambda < \frac{1}{12} \Rightarrow (12\lambda-1) < 0 \therefore V'' > 0 \Rightarrow$ stable | A1 **[2]** | |
---3\\
\includegraphics[max width=\textwidth, alt={}, center]{57323af2-8cf3-4721-b2c8-a968264be343-2_439_444_1318_822}
A uniform rod $A B$ has mass $m$ and length $4 a$. The rod can rotate in a vertical plane about a smooth fixed horizontal axis passing through $A$. One end of a light elastic string of natural length $a$ and modulus of elasticity $\lambda m g$ is attached to $B$. The other end of the string is attached to a small light ring which slides on a fixed smooth horizontal rail which is in the same vertical plane as the rod. The rail is a vertical distance $3 a$ above $A$. The string is always vertical and the rod makes an angle $\theta$ radians with the horizontal, where $0 \leqslant \theta \leqslant \frac { 1 } { 2 } \pi$ (see diagram).\\
(i) Taking $A$ as the reference level for gravitational potential energy, find an expression for the total potential energy $V$ of the system, and show that
$$\frac { \mathrm { d } V } { \mathrm {~d} \theta } = 2 m g a \cos \theta ( 4 \lambda ( 1 + 2 \sin \theta ) - 1 ) .$$
Determine the positions of equilibrium and the nature of their stability in the cases\\
(ii) $\lambda > \frac { 1 } { 12 }$,\\
(iii) $\lambda < \frac { 1 } { 12 }$.\\
\includegraphics[max width=\textwidth, alt={}, center]{57323af2-8cf3-4721-b2c8-a968264be343-3_392_689_269_671}
The diagram shows the curve with equation $y = \frac { 1 } { 2 } \ln x$. The region $R$, shaded in the diagram, is bounded by the curve, the $x$-axis and the line $x = 4$. A uniform solid of revolution is formed by rotating $R$ completely about the $y$-axis to form a solid of volume $V$.\\
(i) Show that $V = \frac { 1 } { 4 } \pi ( 64 \ln 2 - 15 )$.\\
(ii) Find the exact $y$-coordinate of the centre of mass of the solid.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{57323af2-8cf3-4721-b2c8-a968264be343-4_385_741_269_646}
\captionsetup{labelformat=empty}
\caption{Fig. 1}
\end{center}
\end{figure}
Fig. 1 shows part of the line $y = \frac { a } { h } x$, where $a$ and $h$ are constants. The shaded region bounded by the line, the $x$-axis and the line $x = h$ is rotated about the $x$-axis to form a uniform solid cone of base radius $a$, height $h$ and volume $\frac { 1 } { 3 } \pi a ^ { 2 } h$. The mass of the cone is $M$.\\
(i) Show by integration that the moment of inertia of the cone about the $y$-axis is $\frac { 3 } { 20 } M \left( a ^ { 2 } + 4 h ^ { 2 } \right)$. (You may assume the standard formula $\frac { 1 } { 4 } m r ^ { 2 }$ for the moment of inertia of a uniform disc about a diameter.)
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{57323af2-8cf3-4721-b2c8-a968264be343-4_501_556_1238_726}
\captionsetup{labelformat=empty}
\caption{Fig. 2}
\end{center}
\end{figure}
A uniform solid cone has mass 3 kg , base radius 0.4 m and height 1.2 m . The cone can rotate about a fixed vertical axis passing through its centre of mass with the axis of the cone moving in a horizontal plane. The cone is rotating about this vertical axis at an angular speed of $9.6 \mathrm { rad } \mathrm { s } ^ { - 1 }$. A stationary particle of mass $m \mathrm {~kg}$ becomes attached to the vertex of the cone (see Fig. 2). The particle being attached to the cone causes the angular speed to change instantaneously from $9.6 \mathrm { rad } \mathrm { s } ^ { - 1 }$ to $7.8 \mathrm { rad } \mathrm { s } ^ { - 1 }$.\\
(ii) Find the value of $m$.\\
\includegraphics[max width=\textwidth, alt={}, center]{57323af2-8cf3-4721-b2c8-a968264be343-5_534_501_255_767}
A triangular frame $A B C$ consists of three uniform rods $A B , B C$ and $C A$, rigidly joined at $A , B$ and $C$. Each rod has mass $m$ and length $2 a$. The frame is free to rotate in a vertical plane about a fixed horizontal axis passing through $A$. The frame is initially held such that the axis of symmetry through $A$ is vertical and $B C$ is below the level of $A$. The frame starts to rotate with an initial angular speed of $\omega$ and at time $t$ the angle between the axis of symmetry through $A$ and the vertical is $\theta$ (see diagram).\\
(i) Show that the moment of inertia of the frame about the axis through $A$ is $6 m a ^ { 2 }$.\\
(ii) Show that the angular speed $\dot { \theta }$ of the frame when it has turned through an angle $\theta$ satisfies
$$a \dot { \theta } ^ { 2 } = a \omega ^ { 2 } - k g \sqrt { 3 } ( 1 - \cos \theta ) ,$$
stating the exact value of the constant $k$.\\
Hence find, in terms of $a$ and $g$, the set of values of $\omega ^ { 2 }$ for which the frame makes complete revolutions.
At an instant when $\theta = \frac { 1 } { 6 } \pi$, the force acting on the frame at $A$ has magnitude $F$.\\
(iii) Given that $\omega ^ { 2 } = \frac { 2 g } { a \sqrt { 3 } }$, find $F$ in terms of $m$ and $g$.
\section*{END OF QUESTION PAPER}
\hfill \mbox{\textit{OCR M4 2017 Q3 [17]}}