Photo AI
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 ≤
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:
From these equations, we can derive:
- By substituting T from the second equation into the first:
- 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:
- Using the relationship between the angle and radius, where we define it in terms of the natural length h, we derive:
Step 2
(a)(ii) Find the value of θ.
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:
- 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.
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 × 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:
Step 4
(b)(ii) Show that while the string is taut, P is moving with simple harmonic motion.
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:
- This signifies
- As T = kx, leading us to:
- Constant proportionality means this satisfies the criteria of simple harmonic motion.
Step 5