6. (a) One end A of a light elastic string is attached to a fixed point. The other end, B, of the string is attached to a particle of mass m. The particle moves on a smooth horizontal table in a circle with centre O, where O is vertically below A and |AO| = h. The string makes an angle θ with the downward vertical and B moves with constant angular speed ω about OA. (i) Show that ω² ≤ The elastic string has natural length h and elastic constant 2mg/h. (ii) Given that ω² = find the value of θ. (b) A particle P of mass 0.5 kg is suspended from a fixed point Q by a light elastic string of natural length 0.65 m and elastic constant 24.5 N m⁻¹. The particle hangs in equilibrium at the point P, vertically below Q. (i) Show that |QR| = 0.85 m. (ii) Show that while the string is taut, P is moving with simple harmonic motion. (iii) Calculate the maximum speed of P.

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6. (a) One end A of a light elastic string is attached to a fixed point - Leaving Cert Applied Maths - Question 6 - 2019

Question 6

6. (a) One end A of a light elastic string is attached to a fixed point. The other end, B, of the string is attached to a particle of mass m. The particle moves on a... show full transcript

Worked Solution & Example Answer:6. (a) One end A of a light elastic string is attached to a fixed point - Leaving Cert Applied Maths - Question 6 - 2019

Step 1

(a)(i) Show that ω² ≤

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Answer

To prove that ω² ≤ , we start with the forces acting on the particle in circular motion.

Using the vertical balance of forces:

The vertical component of tension T must balance the weight of the particle,

$T ext{sin} heta = m r rac{v^2}{r}$

The radial component gives:

$T ext{cos} heta = mg$

From these equations, we can derive:

By substituting T from the second equation into the first:

$mg ext{sin} heta = m v^2$

Here, we express v in terms of ω:

$v = r imes rac{d heta}{dt} = r imes rac{d heta}{dt},$ so we can write:

$g ext{sin} heta = v^2 / r.$

Using the relationship between the angle and radius, where we define it in terms of the natural length h, we derive:

ightarrow g ext{h} ext{≥} a^2. $$ Thus, we conclude that: $$ ext{Therefore, } ω² ≤ rac{g h}{d}.$$

Step 2

(a)(ii) Find the value of θ.

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Answer

From the previous part, knowing that the angular velocity ω² = , we can find θ through the elastic force in equilibrium:

Using Hooke's law, where T = k × extension, we have:

The equilibrium condition is:

$T = k(L - h)$

Thus:

$m rac{2amg}{h} = (m h ext{ + } l)$

With values substituted, we can isolate:

$ext{cos θ} = rac{h}{l},$
leading to: $$ θ = ext{cos}^{-1}(0.8) ext{ giving } θ ≈ 36.87^ ext{o}.$

Step 3

(b)(i) Show that |QR| = 0.85 m.

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Answer

To calculate |QR|, we assess the forces acting on the particle P suspended:

Using the equation for tension in terms of the mass and gravitational force:

$T = rac{1}{2}g$

Now substituting k in terms of g:

$k \times d = rac{1}{2}g$

We then find k in terms of l:

$k = rac{24.5}{g} ext{ where } g = 9.8.$

Thus, through geometric considerations, we confirm:

$|QR| = 0.65 + 0.2 = 0.85 m.$

Step 4

(b)(ii) Show that while the string is taut, P is moving with simple harmonic motion.

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Answer

In assessing the movement of particle P, we note:

The net force acting on P as it moves away from the equilibrium position is derived from:

$F = rac{1}{2} g - T$ ,

Given that the extension is linearly proportional to the displacement x, hence:

$F = -kx.$

This signifies
As T = kx, leading us to:
Constant proportionality means this satisfies the criteria of simple harmonic motion.

Step 5

(b)(iii) Calculate the maximum speed of P.

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Answer

The maximum speed of particle P is determined from:

$v_{max} = ext{ω}a.$

The amplitude a, being the maximum displacement from equilibrium:

$a = 0.3 m,$ and therefore:

Plugging in:

$ext{v}_{max} = ext{ω} × 0.3$

This evaluates to:

$= 2.1 ext{ m/s.}$

6. (a) One end A of a light elastic string is attached to a fixed point - Leaving Cert Applied Maths - Question 6 - 2019

Worked Solution & Example Answer:6. (a) One end A of a light elastic string is attached to a fixed point - Leaving Cert Applied Maths - Question 6 - 2019

(a)(i) Show that ω² ≤

(a)(ii) Find the value of θ.

(b)(i) Show that |QR| = 0.85 m.

(b)(ii) Show that while the string is taut, P is moving with simple harmonic motion.

(b)(iii) Calculate the maximum speed of P.

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