| Exam Board | OCR |
|---|---|
| Module | M4 (Mechanics 4) |
| Year | 2014 |
| Session | June |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moments of inertia |
| Type | Force at pivot/axis |
| Difficulty | Challenging +1.8 This M4 question requires multiple advanced techniques: calculating moment of inertia using parallel axis theorem, applying energy conservation for rotational motion, and resolving forces in a rotating reference frame. Part (iii) particularly demands careful application of Newton's second law in both radial and tangential directions for a rotating lamina, which is conceptually challenging and involves several coordinated steps beyond standard mechanics problems. |
| 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 formulae6.02i Conservation of energy: mechanical energy principle6.04e 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 |
| MI square centre \(I_c = \frac{1}{3}m(a^2+a^2)\) | B1 | May be implied by later working; B1 \(I_x = \frac{4}{3}ma^2\) |
| \(I_A = \frac{1}{3}m(2a^2) + m(a\sqrt{2})^2 = \frac{8}{3}ma^2\) | M1A1 | M1 for applying parallel axes rule; M1 applying perpendicular axes rule (\(I_x = I_y\)) |
| [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{1}{2}I\omega^2 = mga\sqrt{2}(1-\cos\theta)\) | M1 | Equation involving KE and PE |
| \(\frac{1}{2}\left(\frac{8}{3}ma^2\right)\omega^2 = mga\sqrt{2}(1-\cos\theta)\) | A1 A1 | A1 for KE term, A1 for PE term |
| \(a\omega^2 = \frac{3}{4}g\sqrt{2}(1-\cos\theta)\) | A1 | AG Correctly obtained |
| [4] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(2a\omega\frac{d\omega}{dt} = \frac{3g\sqrt{2}}{4}\sin\theta\frac{d\theta}{dt}\) | M1 | Differentiating \(\omega\) with respect to \(t\); Or M1 for applying \(C = I\alpha\) with their \(I\) from (i) |
| \(\alpha = \frac{3g\sqrt{2}}{8a}\sin\theta\) | A1 | A1 for \(a\sqrt{2}mg\sin\theta = \frac{8}{3}ma^2\alpha\) |
| \(mg\cos\theta - X = ma\sqrt{2}\omega^2\) | M1 | For radial acceleration \(r\omega^2\) |
| \(X = \frac{1}{2}mg(5\cos\theta - 3)\) | A1 | |
| \(mg\sin\theta - Y = ma\sqrt{2}\alpha\) | M1 | For transverse acceleration \(r\alpha\) |
| \(Y = \frac{1}{4}mg\sin\theta\) | A1 | |
| [6] |
# Question 4:
## Part (i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| MI square centre $I_c = \frac{1}{3}m(a^2+a^2)$ | B1 | May be implied by later working; B1 $I_x = \frac{4}{3}ma^2$ |
| $I_A = \frac{1}{3}m(2a^2) + m(a\sqrt{2})^2 = \frac{8}{3}ma^2$ | M1A1 | M1 for applying parallel axes rule; M1 applying perpendicular axes rule ($I_x = I_y$) |
| **[3]** | | |
## Part (ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{1}{2}I\omega^2 = mga\sqrt{2}(1-\cos\theta)$ | M1 | Equation involving KE and PE |
| $\frac{1}{2}\left(\frac{8}{3}ma^2\right)\omega^2 = mga\sqrt{2}(1-\cos\theta)$ | A1 A1 | A1 for KE term, A1 for PE term |
| $a\omega^2 = \frac{3}{4}g\sqrt{2}(1-\cos\theta)$ | A1 | **AG** Correctly obtained |
| **[4]** | | |
## Part (iii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $2a\omega\frac{d\omega}{dt} = \frac{3g\sqrt{2}}{4}\sin\theta\frac{d\theta}{dt}$ | M1 | Differentiating $\omega$ with respect to $t$; Or M1 for applying $C = I\alpha$ with their $I$ from (i) |
| $\alpha = \frac{3g\sqrt{2}}{8a}\sin\theta$ | A1 | A1 for $a\sqrt{2}mg\sin\theta = \frac{8}{3}ma^2\alpha$ |
| $mg\cos\theta - X = ma\sqrt{2}\omega^2$ | M1 | For radial acceleration $r\omega^2$ |
| $X = \frac{1}{2}mg(5\cos\theta - 3)$ | A1 | |
| $mg\sin\theta - Y = ma\sqrt{2}\alpha$ | M1 | For transverse acceleration $r\alpha$ |
| $Y = \frac{1}{4}mg\sin\theta$ | A1 | |
| **[6]** | | |4 A uniform square lamina has mass $m$ and sides of length $2 a$.\\
(i) Calculate the moment of inertia of the lamina about an axis through one of its corners perpendicular to its plane.\\
\includegraphics[max width=\textwidth, alt={}, center]{639c658e-0aca-4161-9e77-0f4c494b0b55-3_693_640_434_715}
The uniform square lamina has centre $C$ and is free to rotate in a vertical plane about a fixed horizontal axis passing through one of its corners $A$. The lamina is initially held such that $A C$ is vertical with $C$ above $A$. The lamina is slightly disturbed from rest from this initial position. When $A C$ makes an angle $\theta$ with the upward vertical, the force exerted by the axis on the lamina has components $X$ parallel to $A C$ and $Y$ perpendicular to $A C$ (see diagram).\\
(ii) Show that the angular speed, $\omega$, of the lamina satisfies $a \omega ^ { 2 } = \frac { 3 } { 4 } g \sqrt { 2 } ( 1 - \cos \theta )$.\\
(iii) Find $X$ and $Y$ in terms of $m , g$ and $\theta$.
\section*{Question 5 begins on page 4.}
\begin{center}
\includegraphics[max width=\textwidth, alt={}]{639c658e-0aca-4161-9e77-0f4c494b0b55-4_767_337_248_863}
\end{center}
A pendulum consists of a uniform rod $A B$ of length $4 a$ and mass $4 m$ and a spherical shell of radius $a$, mass $m$ and centre $C$. The end $B$ of the rod is rigidly attached to a point on the surface of the shell in such a way that $A B C$ is a straight line. The pendulum is initially at rest with $B$ vertically below $A$ and it is free to rotate in a vertical plane about a smooth fixed horizontal axis passing through $A$ (see diagram).\\
(i) Show that the moment of inertia of the pendulum about the axis of rotation is $47 m a ^ { 2 }$.
A particle of mass $m$ is moving horizontally in the plane in which the pendulum is free to rotate. The particle has speed $\sqrt { k g a }$, where $k$ is a positive constant, and strikes the rod at a distance $3 a$ from $A$. In the subsequent motion the particle adheres to the rod and the combined rigid body $P$ starts to rotate.\\
(ii) Show that the initial angular speed of $P$ is $\frac { 3 } { 56 } \sqrt { \frac { k g } { a } }$.\\
(iii) For the case $k = 4$, find the angle that $P$ has turned through when $P$ first comes to instantaneous rest.\\
(iv) Find the least value of $k$ such that the rod reaches the horizontal.\\
\includegraphics[max width=\textwidth, alt={}, center]{639c658e-0aca-4161-9e77-0f4c494b0b55-5_437_903_269_573}
A uniform rod $A B$ has mass $m$ and length $2 a$. 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 $\sqrt { 3 } m g$ is attached to $A$. The string passes over a small smooth fixed pulley $C$, where $A C$ is horizontal and $A C = a$. The other end of the string is attached to the rod at its mid-point $D$. The rod makes an angle $\theta$ below the horizontal (see diagram).\\
(i) Taking $A$ as the reference level for gravitational potential energy, show that the total potential energy $V$ of the system is given by
$$V = m g a ( \sqrt { 3 } - \sin \theta - \sqrt { 3 } \cos \theta ) .$$
(ii) Show that $\theta = \frac { 1 } { 6 } \pi$ is a position of stable equilibrium for the system.
The system is making small oscillations about the equilibrium position.\\
(iii) By differentiating the energy equation with respect to time, show that
$$\frac { 4 } { 3 } a \ddot { \theta } = g ( \cos \theta - \sqrt { 3 } \sin \theta ) .$$
(iv) Using the substitution $\theta = \phi + \frac { 1 } { 6 } \pi$, show that the motion is approximately simple harmonic, and find the approximate period of the oscillations.
\section*{END OF QUESTION PAPER}
\hfill \mbox{\textit{OCR M4 2014 Q4 [13]}}