OCR MEI M4 2012 June — Question 4 25 marks

Exam BoardOCR MEI
ModuleM4 (Mechanics 4)
Year2012
SessionJune
Marks25
PaperDownload PDF ↗
TopicMoments of inertia
TypeProve MI by integration
DifficultyChallenging +1.2 This is a structured multi-part Further Maths mechanics question requiring integration for moment of inertia, application of angular impulse-momentum, and solving a separable differential equation. While it involves several steps and FM content (M4), each part follows standard techniques with clear guidance, making it moderately above average difficulty but not requiring novel insight.
Spec4.03h Determinant 2x2: calculation4.10f Simple harmonic motion: x'' = -omega^2 x6.03e Impulse: by a force6.03f Impulse-momentum: relation6.04a Centre of mass: gravitational effect

4
  1. Show by integration that the moment of inertia of a uniform circular lamina of radius \(a\) and mass \(m\) about an axis perpendicular to the plane of the lamina and through its centre is \(\frac { 1 } { 2 } m a ^ { 2 }\). A closed hollow cylinder has its curved surface and both ends made from the same uniform material. It has mass \(M\), radius \(a\) and height \(h\).
  2. Show that the moment of inertia of the cylinder about its axis of symmetry is \(\frac { 1 } { 2 } M a ^ { 2 } \left( \frac { a + 2 h } { a + h } \right)\). For the rest of this question take the cylinder to have mass 8 kg , radius 0.5 m and height 0.3 m .
    The cylinder is at rest and can rotate freely about its axis of symmetry. It is given a tangential impulse of magnitude 55 Ns at a point on its curved surface. The impulse is perpendicular to the axis.
  3. Find the angular speed of the cylinder after the impulse. A resistive couple is now applied to the cylinder for 5 seconds. The magnitude of the couple is \(2 \dot { \theta } ^ { 2 } \mathrm { Nm }\), where \(\dot { \theta }\) is the angular speed of the cylinder in rad s \({ } ^ { - 1 }\).
  4. Formulate a differential equation for \(\dot { \theta }\) and hence find the angular speed of the cylinder at the end of the 5 seconds. The cylinder is now brought to rest by a constant couple of magnitude 0.03 Nm .
  5. Calculate the time it takes from when this couple is applied for the cylinder to come to rest.

4 (i) Show by integration that the moment of inertia of a uniform circular lamina of radius $a$ and mass $m$ about an axis perpendicular to the plane of the lamina and through its centre is $\frac { 1 } { 2 } m a ^ { 2 }$.

A closed hollow cylinder has its curved surface and both ends made from the same uniform material. It has mass $M$, radius $a$ and height $h$.\\
(ii) Show that the moment of inertia of the cylinder about its axis of symmetry is $\frac { 1 } { 2 } M a ^ { 2 } \left( \frac { a + 2 h } { a + h } \right)$.

For the rest of this question take the cylinder to have mass 8 kg , radius 0.5 m and height 0.3 m .\\
The cylinder is at rest and can rotate freely about its axis of symmetry. It is given a tangential impulse of magnitude 55 Ns at a point on its curved surface. The impulse is perpendicular to the axis.\\
(iii) Find the angular speed of the cylinder after the impulse.

A resistive couple is now applied to the cylinder for 5 seconds. The magnitude of the couple is $2 \dot { \theta } ^ { 2 } \mathrm { Nm }$, where $\dot { \theta }$ is the angular speed of the cylinder in rad s ${ } ^ { - 1 }$.\\
(iv) Formulate a differential equation for $\dot { \theta }$ and hence find the angular speed of the cylinder at the end of the 5 seconds.

The cylinder is now brought to rest by a constant couple of magnitude 0.03 Nm .\\
(v) Calculate the time it takes from when this couple is applied for the cylinder to come to rest.

\hfill \mbox{\textit{OCR MEI M4 2012 Q4 [25]}}