A small bead of mass m is attached to one end of a light string of length R. The other end of the string is fixed at height 2h above the centre of a sphere of radius R, as shown in the diagram. The bead moves in a circle of radius r on the surface of the sphere and has constant angular velocity ω > 0. The string makes an angle of θ with the vertical. There are three forces acting on the bead: the tension force F of the string, the normal reaction force N to the surface of the sphere, and the gravitational force mg. a) (i) By resolving the forces horizontally and vertically on a diagram, show that F sin θ − N sin θ = mω²r. and F cos θ + N cos θ = mg. (ii) Show that N = rac{1}{2} m g sec θ - rac{1}{2} m rac{ω²}{g} r cosec θ. (iii) Show that the bead remains in contact with the sphere if ω ≤ rac{g}{h}. b) If p, q and r are positive real numbers and p + q + r ≥ 0, prove that rac{p}{1 + p} + rac{q}{1 + q} + rac{r}{1 + r} ≥ 0. c) The diagram shows the ellipse rac{x²}{a²} + rac{y²}{b²} = 1, where a > b. The line l is the tangent to the ellipse at the point P. The foci of the ellipse are S and S′. The perpendicular to l through S meets l at the point Q. The lines SQ and S′P meet at the point R. Copy or trace the diagram into your writing booklet. i) Use the reflection property of the ellipse at P to prove that SQ = RQ. ii) Explain why S′R = 2a. iii) Hence, or otherwise, prove that Q lies on the circle x² + y² = a² at the point R.

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A small bead of mass m is attached to one end of a light string of length R - HSC - SSCE Mathematics Extension 2 - Question 5 - 2011 - Paper 1

Question 5

A small bead of mass m is attached to one end of a light string of length R. The other end of the string is fixed at height 2h above the centre of a sphere of radius... show full transcript

Worked Solution & Example Answer:A small bead of mass m is attached to one end of a light string of length R - HSC - SSCE Mathematics Extension 2 - Question 5 - 2011 - Paper 1

Step 1

By resolving the forces horizontally and vertically on a diagram, show that F sin θ − N sin θ = mω²r.

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Answer

To find the horizontal and vertical components of the forces acting on the bead, we start by drawing a diagram indicating the forces involved. For horizontal components, we have:

The horizontal component of the tension: $F \sin{\theta}$
The horizontal component of the normal reaction force: $N \sin{\theta}$

In equilibrium, the net horizontal force is the centripetal force required for circular motion, which gives us:

$F \sin{\theta} − N \sin{\theta} = mω²r$

For the vertical components:

The vertical component of the tension: $F \cos{\theta}$
The normal force: $N$
The weight of the bead: $mg$

The vertical forces balance as: $F \cos{\theta} + N \cos{\theta} = mg$

Step 2

Show that N = \frac{1}{2} m g sec θ - \frac{1}{2} m \frac{ω²}{g} r cosec θ.

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Answer

To isolate N, we can combine the two equations derived earlier. We can substitute for F from one equation into the other:

From the horizontal equation: $F = N \sin{\theta} + mω²r$

Substituting this into the vertical forces equation:

$(N \sin{\theta} + mω²r) \cos{\theta} + N \cos{\theta} = mg$

Rearranging will give: $N(\sin{\theta} \cos{\theta} + \cos{\theta}) = mg - mω²r$

Simplifying this expression gives us:

$N = \frac{mg - mω²r}{\sin{\theta} \cos{\theta} + \cos{\theta}}$

Step 3

Show that the bead remains in contact with the sphere if ω ≤ \frac{g}{h}.

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Answer

To determine the condition under which the bead remains in contact with the sphere, we analyze the derived expression for N. For N to be non-negative (indicating contact), we can rewrite the equation:

$N = \frac{1}{2} m g \sec{θ} - \frac{1}{2} m \frac{ω²}{g} r \csc{θ} ≥ 0$

This leads to: $g \sec{θ} ≥ \frac{ω²}{g} r \csc{θ}$

From here, we simplify further to show that: $ω² ≤ \frac{g^2}{h}$

Thus, if the condition $ω ≤ \frac{g}{h}$ holds, the bead will remain in contact with the sphere.

Step 4

Using a common denominator or multiplying both sides by (1+p)(1+q)(1+r) then correctly simplifying the denominator to prove that \frac{p}{1 + p} + \frac{q}{1 + q} + \frac{r}{1 + r} ≥ 0.

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Answer

We first combine the fractions on the left by obtaining a common denominator:

$\frac{p(1 + q)(1 + r) + q(1 + p)(1 + r) + r(1 + p)(1 + q)}{(1 + p)(1 + q)(1 + r)}$

Now, simplifying the numerator will yield a polynomial in p, q, r that reflects their positivity since p, q, and r are positive reals. Therefore, we can conclude that the entire expression remains non-negative.

Step 5

Use the reflection property of the ellipse at P to prove that SQ = RQ.

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The reflection property of the ellipse states that the angle of incidence equals the angle of reflection. Since the line l is tangent at point P, the angle at which the light rays toward S reflect off the tangent will cause them to intersect at Q. Thus, we can deduce that:

$SQ = RQ$ This derives from the properties of the tangent and the reflective angles.

Step 6

Explain why S′R = 2a.

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By definition, the foci S and S' of an ellipse are located at a distance of c from the center O, where:

$c = \sqrt{a^2 - b^2}$

The distance between the foci S and S', which is $2c = S'S$ , gives us the resulting length along the major axis, satisfying that:

$S′R = 2a$

Step 7

Hence, or otherwise, prove that Q lies on the circle x² + y² = a² at the point R.

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Since Q is the point of intersection derived from the reflection properties and the ellipse's tangent, we can analyze the coordinates of Q that lies along the extension of the major axis. Given the geometric properties laid out, we can show that: $x^2 + y^2 = a^2$ This confirms that point Q lies indeed on the specified circle.

A small bead of mass m is attached to one end of a light string of length R - HSC - SSCE Mathematics Extension 2 - Question 5 - 2011 - Paper 1

Worked Solution & Example Answer:A small bead of mass m is attached to one end of a light string of length R - HSC - SSCE Mathematics Extension 2 - Question 5 - 2011 - Paper 1

By resolving the forces horizontally and vertically on a diagram, show that F sin θ − N sin θ = mω²r.

Show that N = \frac{1}{2} m g sec θ - \frac{1}{2} m \frac{ω²}{g} r cosec θ.

Show that the bead remains in contact with the sphere if ω ≤ \frac{g}{h}.

Using a common denominator or multiplying both sides by (1+p)(1+q)(1+r) then correctly simplifying the denominator to prove that \frac{p}{1 + p} + \frac{q}{1 + q} + \frac{r}{1 + r} ≥ 0.

Use the reflection property of the ellipse at P to prove that SQ = RQ.

Explain why S′R = 2a.

Hence, or otherwise, prove that Q lies on the circle x² + y² = a² at the point R.

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