OCR M4 2004 January — Question 6 13 marks

Exam BoardOCR
ModuleM4 (Mechanics 4)
Year2004
SessionJanuary
Marks13
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicMoments of inertia
TypeForce at pivot/axis
DifficultyStandard +0.3 This is a standard M4 rotation question with straightforward applications of center of mass, moment of inertia formulas, and rotational dynamics equations. Parts (i)-(iii) involve routine calculations using standard formulas, while part (iv) requires resolving forces but follows a predictable method. The question is slightly above average difficulty due to the multi-step nature and the need to coordinate several concepts, but all techniques are standard textbook exercises for M4 students.
Spec6.04d Integration: for centre of mass of laminas/solids6.05f Vertical circle: motion including free fall

6 A rigid body consists of a uniform rod \(A B\), of mass 15 kg and length 2.8 m , with a particle of mass 5 kg attached at \(B\). The body rotates without resistance in a vertical plane about a fixed horizontal axis through \(A\).
  1. Find the distance of the centre of mass of the body from \(A\).
  2. Find the moment of inertia of the body about the axis. \includegraphics[max width=\textwidth, alt={}, center]{4cac1898-8251-4cda-bbbc-c2c30fde5a6e-3_475_682_680_719} At one instant, \(A B\) makes an acute angle \(\theta\) with the downward vertical, the angular speed of the body is \(1.2 \mathrm { rad } \mathrm { s } ^ { - 1 }\) and the angular acceleration of the body is \(3.5 \mathrm { rad } \mathrm { s } ^ { - 2 }\) (see diagram).
  3. Show that \(\sin \theta = 0.8\).
  4. Find the components, parallel and perpendicular to \(B A\), of the force acting on the body at \(A\).
    [0pt] [Question 7 is printed overleaf.] \includegraphics[max width=\textwidth, alt={}, center]{4cac1898-8251-4cda-bbbc-c2c30fde5a6e-4_949_1112_281_550} A small bead \(B\), of mass \(m\), slides on a smooth circular hoop of radius \(a\) and centre \(O\) which is fixed in a vertical plane. A light elastic string has natural length \(2 a\) and modulus of elasticity \(m g\); one end is attached to \(B\), and the other end is attached to a light ring \(R\) which slides along a smooth horizontal wire. The wire is in the same vertical plane as the hoop, and at a distance \(2 a\) above \(O\). The elastic string \(B R\) is always vertical, and \(O B\) makes an angle \(\theta\) with the downward vertical (see diagram).
  5. Show that \(\theta = 0\) is a position of stable equilibrium.
  6. Find the approximate period of small oscillations about the equilibrium position \(\theta = 0\).

\(M\bar{x} = \sum m_i x_i\)
\(20\bar{x} = 15 \times 1.4 + 5 \times 2.8\)
AnswerMarks Guidance
\(\bar{x} = 1.75\)M1 A1 [2] Centre of mass calculation
\(I = \frac{4}{3}ml^2 + 5 \cdot 2l^2 = \frac{4}{3} \times 15 \times 1.4^2 + 20 \times 1.4^2 = 78.4\) kg m²M1 A1 [2] Moment of inertia about pivot
\(C = I\alpha\)
\(196 \times 1.75\sin\theta = 78.4 \times 3.5\)
AnswerMarks Guidance
\(\sin\theta = 0.8\) (show)M1 A1 [3] Torque equation; verification that \(\sin\theta = 0.8\)
N2 radially BA:
\(H - mg\cos\theta = m\omega^2 r\)
AnswerMarks Guidance
\(H = 20 \times 9.8 \times 0.6 + 20 \times 1.2^2 \times 1.75 = 168\) NM1 A1 Radial component of equation; correct substitution
N2 normal:
\(mg\sin\theta - P = mr\omega\)
AnswerMarks Guidance
\(P = 20 \times 9.8 \times 0.8 - 20 \times 1.75 \times 3.5 = 34.3\)M1 A1 [6] Tangential component; correct substitution
$M\bar{x} = \sum m_i x_i$
$20\bar{x} = 15 \times 1.4 + 5 \times 2.8$
$\bar{x} = 1.75$ | M1 A1 | [2] Centre of mass calculation

$I = \frac{4}{3}ml^2 + 5 \cdot 2l^2 = \frac{4}{3} \times 15 \times 1.4^2 + 20 \times 1.4^2 = 78.4$ kg m² | M1 A1 | [2] Moment of inertia about pivot

$C = I\alpha$
$196 \times 1.75\sin\theta = 78.4 \times 3.5$
$\sin\theta = 0.8$ (show) | M1 A1 | [3] Torque equation; verification that $\sin\theta = 0.8$

**N2 radially BA:**
$H - mg\cos\theta = m\omega^2 r$
$H = 20 \times 9.8 \times 0.6 + 20 \times 1.2^2 \times 1.75 = 168$ N | M1 A1 | Radial component of equation; correct substitution

**N2 normal:**
$mg\sin\theta - P = mr\omega$
$P = 20 \times 9.8 \times 0.8 - 20 \times 1.75 \times 3.5 = 34.3$ | M1 A1 | [6] Tangential component; correct substitution

---
6 A rigid body consists of a uniform rod $A B$, of mass 15 kg and length 2.8 m , with a particle of mass 5 kg attached at $B$. The body rotates without resistance in a vertical plane about a fixed horizontal axis through $A$.\\
(i) Find the distance of the centre of mass of the body from $A$.\\
(ii) Find the moment of inertia of the body about the axis.\\
\includegraphics[max width=\textwidth, alt={}, center]{4cac1898-8251-4cda-bbbc-c2c30fde5a6e-3_475_682_680_719}

At one instant, $A B$ makes an acute angle $\theta$ with the downward vertical, the angular speed of the body is $1.2 \mathrm { rad } \mathrm { s } ^ { - 1 }$ and the angular acceleration of the body is $3.5 \mathrm { rad } \mathrm { s } ^ { - 2 }$ (see diagram).\\
(iii) Show that $\sin \theta = 0.8$.\\
(iv) Find the components, parallel and perpendicular to $B A$, of the force acting on the body at $A$.\\[0pt]
[Question 7 is printed overleaf.]\\
\includegraphics[max width=\textwidth, alt={}, center]{4cac1898-8251-4cda-bbbc-c2c30fde5a6e-4_949_1112_281_550}

A small bead $B$, of mass $m$, slides on a smooth circular hoop of radius $a$ and centre $O$ which is fixed in a vertical plane. A light elastic string has natural length $2 a$ and modulus of elasticity $m g$; one end is attached to $B$, and the other end is attached to a light ring $R$ which slides along a smooth horizontal wire. The wire is in the same vertical plane as the hoop, and at a distance $2 a$ above $O$. The elastic string $B R$ is always vertical, and $O B$ makes an angle $\theta$ with the downward vertical (see diagram).\\
(i) Show that $\theta = 0$ is a position of stable equilibrium.\\
(ii) Find the approximate period of small oscillations about the equilibrium position $\theta = 0$.

\hfill \mbox{\textit{OCR M4 2004 Q6 [13]}}