Answer:
Tension: 0.95 N
The particle can't complete it's circular path.
Explanation:
let's answer this question step by step:
Finding Tension in the String:
At the given position, the tension [tex]\sf (T)[/tex] in the string is the sum of two forces:
1) Centripetal force: This force acts inwards, pulling the particle towards the center of the circle. It's calculated as:
[tex]\sf \textsf{Centripetal force} = \dfrac{m \cdot v^2}{r} [/tex]
where:
- [tex]\sf m[/tex] = mass of the particle (0.1 kg),
- [tex]\sf v[/tex] = instantaneous speed (1 m/s),
- [tex]\sf r[/tex] = radius of the circle (1 m).
2) Gravitational force: This force acts downwards due to gravity. Its component along the string is:
[tex]\sf \textsf{Gravitational force component} = m \cdot g \cdot \cos(\theta) [/tex]
where:
- [tex]\sf g[/tex] = acceleration due to gravity (approximately 9.81 m/s²),
- [tex]\sf \theta[/tex] = angle between the string and the vertical (30°).
Therefore, the total tension:
[tex]\sf T = \textsf{Centripetal force} + \textsf{Gravitational force component} [/tex]
[tex]\sf T = \dfrac{m \cdot v^2}{r} + m \cdot g \cdot \cos(\theta) [/tex]
Substitute the given value:
[tex]\sf T = \dfrac{0.1 \cdot 1^2}{1} + 0.1 \cdot 9.81 \cdot \cos(30°) [/tex]
Simplify further:
[tex]\sf T = 0.1 + 0.8495709211125 [/tex]
[tex]\sf T \approx 0.9495709211125 [/tex]
[tex]\sf T \approx 0.95 \, \textsf{N (in 2 d.p.)} [/tex]
2. Can the Particle Complete the Circle?
The minimum speed required for the particle to complete the circle is:
[tex]\sf \textsf{Minimum speed} = \sqrt{5 \cdot g \cdot r} [/tex]
[tex]\sf \textsf{Minimum speed} = \sqrt{5 \cdot 9.81 \cdot 1} [/tex]
[tex]\sf \textsf{Minimum speed} \approx 7.0035705179572 \, \textsf{m/s} [/tex]
[tex]\sf \textsf{Minimum speed} \approx 7 \, \textsf{m/s (in nearest whole number)} [/tex]
Now finding the initial speed:
By applying conservation of energy at two points, we get
[tex]\sf \dfrac{1}{2}mu^{2}=\dfrac{1}{2}m(1)^{2}+mgR(1-cos(30^{\circ})[/tex]
[tex]\sf u^{2}=1+2g \times 1(1-\dfrac{\sqrt{3}}{2})[/tex] [tex]\sf [tex]\sf u^2 =1+2\times 9.81 (1-\dfrac{\sqrt{3}}{2})[/tex]
[tex]\sf u^2 = 3.628581578 [/tex]
[tex]\sf u =\sqrt{3.628581578}[/tex]
[tex]\sf u = 1.904883613 [/tex]
[tex]\sf u = 1.9 \textsf{m/s (in nearest tenth)}[/tex]
So, the instantaneous speed (1.9 m/s) is less than the minimum speed, the particle cannot complete its circular path. It will lose its upward momentum and eventually fall below the starting point.
Therefore, the tension in the string at the given position is approximately 0.95 N, and the particle cannot complete the circular path due to insufficient speed.
