Exam-Style Problem

9231 P21 - Nov 2017 - Q11E - 14 marks

6174

Question 11 EITHER alternative.

A particle \(P\) of mass \(m\) is free to move on the smooth inner surface of a fixed hollow sphere of radius \(a\). The centre of the sphere is \(O\). The points \(A\) and \(A'\) are on the inner surface of the sphere, on opposite sides of the vertical through \(O\); the radius \(OA\) makes an angle \(\alpha\) with the downward vertical and the radius \(OA'\) makes an angle \(\beta\) with the upward vertical. The point \(B\) is on the inner surface of the sphere, vertically below \(O\). The point \(B'\) is on the inner surface of the sphere and such that \(OB'\) makes an angle \(2\beta\) with the upward vertical through \(O\). It is given that \(\cos\alpha=\frac{1}{16}\).

(i) \(P\) is projected from \(A\) with speed \(u\) along the surface of the sphere downwards towards \(B\). Subsequently it loses contact with the sphere at \(A'\). Show that \(u^2=\frac18ag(1+24\cos\beta)\).

(ii) \(P\) is now projected from \(B\) with speed \(u\) along the surface of the sphere towards \(B'\). Subsequently it loses contact with the sphere at \(B'\). Find \(\cos\beta\).

(iii) In part (i), the reaction of the sphere on \(P\) when it is initially projected at \(A\) is \(R\). Find \(R\) in terms of \(m\) and \(g\).

Solution Checked by expert

Answer: \(u^2=\dfrac18ag(1+24\cos\beta)\), \(\boxed{\cos\beta=\dfrac34}\), and \(\boxed{R=\dfrac{39}{16}mg}\).

(i) Let the speed of \(P\) at \(A'\) be \(v\).

From \(A\) to \(A'\), the particle rises by a vertical distance

\[a\cos\alpha+a\cos\beta.\]

Using conservation of energy,

\[\frac12mv^2=\frac12mu^2-mga(\cos\alpha+\cos\beta).\]

At \(A'\), the particle is just losing contact with the sphere, so the reaction is zero.

The radial equation towards the centre is therefore

\[\frac{mv^2}{a}=mg\cos\beta.\]

Hence

\[v^2=ag\cos\beta.\]

From the energy equation,

\[u^2=v^2+2ag(\cos\alpha+\cos\beta).\]

Substitute \(v^2=ag\cos\beta\) and \(\cos\alpha=\frac{1}{16}\):

\[u^2=ag\cos\beta+2ag\left(\frac{1}{16}+\cos\beta\right).\]

\[u^2=ag\left(3\cos\beta+\frac18\right).\]

Therefore

\[u^2=\frac18ag(1+24\cos\beta),\]

as required.

(ii) Now the particle is projected from \(B\) and loses contact at \(B'\).

Let its speed at \(B'\) be \(w\). From \(B\) to \(B'\), the rise in height is

\[a(1+\cos2\beta).\]

Conservation of energy gives

\[\frac12mw^2=\frac12mu^2-mga(1+\cos2\beta).\]

At \(B'\), the reaction is zero, so the radial equation gives

\[\frac{mw^2}{a}=mg\cos2\beta.\]

Thus

\[w^2=ag\cos2\beta.\]

\[u^2=w^2+2ag(1+\cos2\beta).\]

Hence

\[u^2=ag\cos2\beta+2ag(1+\cos2\beta)=ag(2+3\cos2\beta).\]

But from part (i),

\[u^2=\frac18ag(1+24\cos\beta).\]

Equate the two expressions and cancel \(ag\):

\[\frac18(1+24\cos\beta)=2+3\cos2\beta.\]

Multiply by \(8\):

\[1+24\cos\beta=16+24\cos2\beta.\]

Use \(\cos2\beta=2\cos^2\beta-1\):

\[1+24c=16+24(2c^2-1),\]

where \(c=\cos\beta\).

\[1+24c=48c^2-8.\]

Rearranging gives

\[48c^2-24c-9=0.\]

Divide by \(3\):

\[16c^2-8c-3=0.\]

Factorising,

\[(4c-3)(4c+1)=0.\]

Thus

\[c=\frac34\quad\text{or}\quad c=-\frac14.\]

The valid value here is

\[\boxed{\cos\beta=\frac34}.\]

(iii) Using \(\cos\beta=\frac34\) in the expression from part (i),

\[u^2=\frac18ag\left(1+24\cdot\frac34\right).\]

\[u^2=\frac18ag(1+18)=\frac{19}{8}ag.\]

At \(A\), the radial equation towards the centre gives

\[R=\frac{mu^2}{a}+mg\cos\alpha.\]

Substitute \(u^2=\frac{19}{8}ag\) and \(\cos\alpha=\frac{1}{16}\):

\[R=m\cdot\frac{19g}{8}+mg\cdot\frac{1}{16}.\]

Therefore

\[R=\left(\frac{38}{16}+\frac{1}{16}\right)mg=\boxed{\frac{39}{16}mg}.\]