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STEP II 2007 question 10 solution

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TSR Wiki > Study Help > Subjects and Revision > Mathematics > STEP > STEP II 2007 question 10 solution

Image:II07Q10.gif

Let X make an angle \alpha to the horizontal plane and let Y be perpendicular to X. By symmetry, \bar{X} = r

The mass of the cylinder is 3\pi r^{3} \rho. The mass of the hemisphere is 2\pi r^{3} \rho

The centre of mass of the cylinder is (r, 1.5r). The centre of mass of the hemisphere is (r, 3.375r)

\bar{Y} = \frac{4.5r + 6.75r}{5} = 2.25r

Resolving vertically: R = 5\pi r^{3} \rho Resolving horizontally: F = P \Rightarrow 5\pi r^{3} \rho \mu \geq P

Before we take moments about the point of the cylinder in contact with the ground (let's call it point A), it is necessary to find the distances involved.

By referring to the diagram, note that the horizontal distance between A and the weight is 2.25r\sin\alpha - r\cos\alpha.

Also, the vertical distance between A and P is 2r\sin\alpha.

Taking moments about A: 5\pi r^{4} \rho (2.25\sin\alpha - \cos\alpha) = 2r\sin\alpha P

\Rightarrow P = \frac{5\pi r^{3} \rho (2.25\sin\alpha - \cos\alpha)}{2\sin\alpha} \leq 5\pi r^{3} \rho \mu

\Rightarrow \mu \geq \frac{2.25\sin\alpha - \cos\alpha}{2\sin\alpha} = \frac{9}{8} - \frac{1}{2}\cot\alpha

When this expression is negative, the force P acts in the opposite direction to that shown in the diagram. F must also act in the opposite direction.

The magnitude of P is \frac{5\pi r^{3} \rho (\cos\alpha-2.25\sin\alpha) }{2\sin\alpha} \leq 5\pi r^{3} \rho \mu

\Rightarrow \mu \geq | \frac{9}{8} - \frac{1}{2}\cot\alpha |

Solution by Dystopia.

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