Exam-Style Problem

9231 P23 - Jun 2017 - Q11E - 12 marks

6162

Question 11 EITHER alternative.

The diagram shows a uniform thin \(\operatorname{rod} A B\) of length \(3 a\) and mass \(8 m\). The end \(A\) is rigidly attached to the surface of a sphere with centre \(O\) and radius \(a\). The rod is perpendicular to the surface of the sphere. The sphere consists of two parts: an inner uniform solid sphere of mass \(\frac{3}{2} m\) and radius \(a\) surrounded by a thin uniform spherical shell of mass \(m\) and also of radius \(a\). The horizontal axis \(l\) is perpendicular to the rod and passes through the point \(C\) on the rod where \(A C=a\).
(i) Show that the moment of inertia of the object, consisting of rod, shell and inner sphere, about the axis \(l\) is \(\frac{289}{15} m a^{2}\).

The object is free to rotate about the axis \(l\). The object is held so that \(C A\) makes an angle \(\alpha\) with the downward vertical and is released from rest.
(ii) Given that \(\cos \alpha=\frac{1}{6}\), find the greatest speed achieved by the centre of the sphere in the subsequent motion.

Solution Checked by expert

Answer: \(I=\dfrac{289}{15}ma^2\), and the greatest speed of the centre of the sphere is \(\boxed{\dfrac{10}{17}\sqrt{ag}}\).

(i) First find the moment of inertia of each part about the axis \(l\).

The rod has mass \(8m\) and length \(3a\). Its centre is \(\frac{3a}{2}\) from \(A\), while \(C\) is \(a\) from \(A\). So the centre of the rod is \(\frac{a}{2}\) from the axis.

Hence, using the parallel-axis theorem,

\[I_{AB}=\frac{1}{12}(8m)(3a)^2+8m\left(\frac a2\right)^2.\]

\[I_{AB}=6ma^2+2ma^2=8ma^2.\]

The centre of the sphere is \(2a\) from \(C\). For the thin spherical shell,

\[I_{\text{shell}}=\frac23ma^2+m(2a)^2.\]

Thus

\[I_{\text{shell}}=\frac23ma^2+4ma^2=\frac{14}{3}ma^2.\]

For the inner solid sphere, its mass is \(\frac32m\), so

\[I_{\text{sphere}}=\frac25\left(\frac32m\right)a^2+\left(\frac32m\right)(2a)^2.\]

Therefore

\[I_{\text{sphere}}=\frac35ma^2+6ma^2=\frac{33}{5}ma^2.\]

Now add the three moments of inertia:

\[I=8ma^2+\frac{14}{3}ma^2+\frac{33}{5}ma^2.\]

Using a common denominator of 15,

\[I=\left(\frac{120}{15}+\frac{70}{15}+\frac{99}{15}\right)ma^2=\frac{289}{15}ma^2.\]

This proves the required result.

(ii) The greatest speed occurs when \(CA\) is vertical, because this is the position of lowest gravitational potential energy.

The sphere, including the shell and inner solid sphere, has total mass

\[m+\frac32m=\frac52m.\]

Its centre is \(2a\) from \(C\), so as \(CA\) moves to the downward vertical, its vertical fall is

\[2a(1-\cos\alpha).\]

The rod's centre is \(\frac a2\) on the other side of \(C\), so it rises by

\[\frac a2(1-\cos\alpha).\]

Therefore the net loss of gravitational potential energy is

\[\left(\frac52mg\right)2a(1-\cos\alpha)-8mg\left(\frac a2\right)(1-\cos\alpha).\]

This simplifies to

\[mga(1-\cos\alpha).\]

Using \(\cos\alpha=\frac16\), the loss of potential energy is

\[mga\left(1-\frac16\right)=\frac56mga.\]

This becomes rotational kinetic energy:

\[\frac12I\omega^2=\frac56mga.\]

Substitute \(I=\frac{289}{15}ma^2\):

\[\frac12\cdot\frac{289}{15}ma^2\omega^2=\frac56mga.\]

Solving gives

\[\omega^2=\frac{25g}{289a}.\]

The centre of the sphere is \(2a\) from the axis, so its speed is

\[v=2a\omega.\]

Hence

\[v=2a\sqrt{\frac{25g}{289a}}=\boxed{\frac{10}{17}\sqrt{ag}}.\]