M3 January 2013 Q6
6.
A smooth hollow cylinder of internal radius \(a\) is fixed with its axis horizontal. A particle \(P\) moves on the inner surface of the cylinder in a vertical circle with radius \(a\) and centre \(O\), where \(O\) lies on the axis of the cylinder. The particle is projected vertically downwards with speed \(u\) from point \(A\) on the circle, where \(OA\) is horizontal. The particle first loses contact with the cylinder at the point \(B\), where \(\angle AOB = 150^\circ\), as shown in Figure 3. Given that air resistance can be ignored,
(a) show that the speed of \(P\) at \(B\) is \(\sqrt{\left(\dfrac{ag}{2}\right)}\), (3)
(b) find \(u\) in terms of \(a\) and \(g\). (4)
After losing contact with the cylinder, \(P\) crosses the diameter through \(A\) at the point \(D\). At \(D\) the velocity of \(P\) makes an angle \(\theta^\circ\) with the horizontal.
(c) Find the value of \(\theta\). (7)
| Scheme | Marks |
|---|---|
| At \(B\) \(mg\cos 60\ \ (+R) = \dfrac{mv^2}{a}\) | M1A1 |
| \(\dfrac{1}{2}g = \dfrac{v^2}{a}\ \ \ \ \ \ v = \sqrt{\dfrac{ag}{2}}\) * | A1 |
| Scheme | Marks |
|---|---|
| Energy \(A\) to \(B\): \(\dfrac{1}{2}mu^2 - \dfrac{1}{2}m\left(\dfrac{ag}{2}\right) = mga\sin 30\) | M1A1A1 |
| \(u^2 = \dfrac{ag}{2} + 2ag \times \dfrac{1}{2}\) | |
| \(u = \sqrt{\dfrac{3ag}{2}}\) | A1 |
| Scheme | Marks |
|---|---|
| Horiz speed \(= \sqrt{\dfrac{ag}{2}}\cos 60\ \ \ \left(= \dfrac{1}{2}\sqrt{\dfrac{ag}{2}}\right)\) | M1A1 |
| Initial vert speed \(= (-)\sqrt{\dfrac{ag}{2}}\sin 60\ \ \ \left(= (-)\dfrac{1}{2}\sqrt{\dfrac{3ag}{2}}\right)\) | M1 |
| \(v^2 = \dfrac{1}{4} \times \dfrac{3ag}{2} + 2g \times \dfrac{a}{2}\) | M1A1 |
| \(v^2 = \dfrac{11ag}{8}\) | |
| \(\tan\theta = \dfrac{\text{vert}}{\text{horiz}} = \sqrt{\dfrac{11ag}{8} \times \dfrac{8}{ag}} = \sqrt{11}\) | M1 |
| \(\theta = 73.22\ldots = 73\) | A1 |