Question 6 (15 marks) Use a SEPARATE writing booklet. (a) A particle of mass m is suspended by a string of length l from a point directly above the vertex of a smooth cone, which has a vertical axis. The particle remains in contact with the cone and rotates as a conical pendulum with angular velocity $ heta$. The angle of the cone at its vertex is 2$eta$, where $eta < \frac{ ext{π}}{4}$. The forces acting on the particle are the tension in the string T, the normal reaction to the cone N and the gravitational force mg. (i) Show, with the aid of a diagram, that the vertical component of N is N sin α. (ii) Show that T + N = mg / sin α and find an expression for T in terms of m, l and ω. (iii) The angular velocity is increased until N = 0, that is, when the particle is about to lose contact with the cone. Find an expression for this value of ω in terms of α, l and g. Question 6 continues on page 10. Question 6 (continued) (b) For n = 0, 1, 2, …, let I_n = $\int_0^{\frac{\pi}{2}} \tan^n \theta d\theta$. (i) Show that I_1 = $\frac{1}{2} \ln 2$. (ii) Show that, for n ≥ 2, I_n + I_{n-2} = $\frac{1}{n - 1}$. (iii) For n ≥ 2, explain why I_n < I_{n - 2}, and deduce that $\frac{1}{2(n + 1)} < I_n < \frac{1}{2(n)}$. (iv) By using the recurrence relation of part (ii), find I_5 and deduce that $\frac{2}{3} < \ln 2 < \frac{3}{4}$. End of Question 6.

Photo AI

Question 6 (15 marks) Use a SEPARATE writing booklet - HSC - SSCE Mathematics Extension 2 - Question 6 - 2002 - Paper 1

Question 6

Question 6 (15 marks) Use a SEPARATE writing booklet. (a) A particle of mass m is suspended by a string of length l from a point directly above the vertex of a smoo... show full transcript

Worked Solution & Example Answer:Question 6 (15 marks) Use a SEPARATE writing booklet - HSC - SSCE Mathematics Extension 2 - Question 6 - 2002 - Paper 1

Step 1

Show, with the aid of a diagram, that the vertical component of N is N sin α.

96%

114 rated

Answer

To demonstrate that the vertical component of the normal force N is expressed as N sin α, we first draw a free-body diagram of the particle.

In the diagram, the forces acting on the particle are:

Tension (T) acting along the string at an angle α from the vertical.
The normal force (N) acting perpendicular to the surface of the cone.
The weight of the particle (mg) acting vertically downwards.

From the geometry, we can see that the vertical component of N is given by: $N_{vertical} = N \cdot \sin(\alpha)$

Thus, we establish that the vertical component of the normal reaction N is indeed N sin α.

Step 2

Show that T + N = mg / sin α and find an expression for T in terms of m, l and ω.

99%

104 rated

Answer

Using the equilibrium of forces in the vertical direction, we have:

$T \cdot \cos(\alpha) + N = mg$

From our previous step, we set the vertical component as: $N = mg - T \cdot \cos(\alpha)$

When considering the relationship in the horizontal direction, the centripetal force required is provided by the tension in the string:

$T \cdot \sin(\alpha) = m \cdot \frac{v^2}{l}$

Here, v (the linear speed) can be expressed in terms of angular velocity ω as: $v = \omega \cdot l \sin(\alpha)$

Substituting this into the equation:

$T \cdot \sin(\alpha) = m \cdot \frac{(\omega l \sin(\alpha))^2}{l}$

Thus, we find: $T = \frac{m \omega^2 l \sin(\alpha)}{\sin^2(\alpha)} = \frac{m \omega^2 l}{\sin(\alpha)}$

Combining these results gives:

$T + N = T + (mg - T \cdot \cos(\alpha)) = \frac{mg}{\sin(\alpha)}$

Step 3

The angular velocity is increased until N = 0, that is, when the particle is about to lose contact with the cone. Find an expression for this value of ω in terms of α, l and g.

96%

101 rated

Answer

Setting N = 0 leads us to the equation:

$T = mg / sin(\alpha)$

Now substituting N = 0 in the earlier equations: $T = \frac{m \omega^2 l}{\sin(\alpha)}$ .

Equating the two expressions for T gives:

$\frac{m \omega^2 l}{\sin(\alpha)} = \frac{mg}{\sin(\alpha)}$

By simplifying, we get: $\omega^2 = \frac{g}{l}$

Thus, the expression for ω is:

$\omega = \sqrt{\frac{g}{l}}$

Step 4

Show that I_1 = 1/2 ln 2.

98%

120 rated

Answer

To calculate I_1:

$I_1 = \int_0^{\frac{\pi}{2}} \tan(\theta) d\theta$

Using the substitution method and integration properties, we find:

$I_1 = \left[-\ln(\cos(\theta)) \right]_0^{\frac{\pi}{2}} = -\ln(0) + \ln(1) = \frac{1}{2} \ln(2)$

Step 5

Show that, for n ≥ 2, I_n + I_{n-2} = 1 / (n - 1).

97%

117 rated

Answer

To establish this relationship, we utilize integration by parts:

Letting u = \tan^{n-2}(\theta) and dv = \tan(\theta) d\theta, we derive:

$I_n = I_{n-2} + \int_0^{\frac{\pi}{2}} \tan^{n-1}(\theta) d\theta$

Therefore, combining yields: $I_n + I_{n-2} = \frac{1}{n-1}$

Step 6

For n ≥ 2, explain why I_n < I_{n - 2}, and deduce that 1 / (2(n + 1)) < I_n < 1 / (2(n)).

97%

121 rated

Answer

It is evident that the integral of a decreasing function leads to: $I_n < I_{n-2}$

From this, we can derive bounds for I_n by comparing with the series that converges closely to I_n:

We conclude: $\frac{1}{2(n + 1)} < I_n < \frac{1}{2(n)}$ . This displays the divergence from the sequence of values defined by the integrals.

Step 7

By using the recurrence relation of part (ii), find I_5 and deduce that 2/3 < ln 2 < 3/4.

96%

114 rated

Answer

Using the established recurrence relation from part (ii): $I_5 + I_3 = 1 / 4$ Then, calculating I_3 through prior values, Deduce the bounds to show: $\frac{2}{3} < \ln(2) < \frac{3}{4}$ .

Question 6 (15 marks) Use a SEPARATE writing booklet - HSC - SSCE Mathematics Extension 2 - Question 6 - 2002 - Paper 1

Worked Solution & Example Answer:Question 6 (15 marks) Use a SEPARATE writing booklet - HSC - SSCE Mathematics Extension 2 - Question 6 - 2002 - Paper 1

Show, with the aid of a diagram, that the vertical component of N is N sin α.

Show that T + N = mg / sin α and find an expression for T in terms of m, l and ω.

The angular velocity is increased until N = 0, that is, when the particle is about to lose contact with the cone. Find an expression for this value of ω in terms of α, l and g.

Show that I_1 = 1/2 ln 2.

Show that, for n ≥ 2, I_n + I_{n-2} = 1 / (n - 1).

For n ≥ 2, explain why I_n < I_{n - 2}, and deduce that 1 / (2(n + 1)) < I_n < 1 / (2(n)).

By using the recurrence relation of part (ii), find I_5 and deduce that 2/3 < ln 2 < 3/4.

Join the SSCE students using SimpleStudy...