Answer:
[tex] 6.3 [/tex] m/s
Explanation:
Given:
- Diameter of the circular platform, [tex] d = 9.89 [/tex] m
- Mass of each seat, [tex] m = 10 [/tex] kg
- Length of the chains, [tex] l = 1.72 [/tex] m
- Angle the chains make with the vertical, [tex] \theta = 34.4^\circ [/tex]
- Acceleration due to gravity, [tex] g = 9.8 [/tex] m/s[tex]^2[/tex]
We need to find the speed of each seat [tex] v [/tex].
We'll start by finding the radius [tex] r [/tex] of the circular platform. The radius can be calculated using the formula:
[tex] r = \dfrac{d}{2} + l \sin(\theta) [/tex]
Let's substitute the values:
[tex] r = \dfrac{9.89}{2} + 1.72 \times \sin(34.4^\circ) [/tex]
[tex] r = 4.945 + 1.72 \times 0.573 [/tex]
[tex] r = 4.945 + 0.986 [/tex]
[tex] r = 5.931 \, \textsf{m} [/tex]
Now, let's calculate the tension in the chain. From the given conditions, the vertical component of the tension balances the weight of the seat, and the horizontal component provides the centripetal force.
Using the vertical balance of forces:
[tex] T \cos(\theta) = mg [/tex]
[tex] T = \dfrac{mg}{\cos(\theta)} [/tex]
Using the horizontal component of the tension to provide the centripetal force:
[tex] T \sin(\theta) = \dfrac{mv^2}{r} [/tex]
We can substitute the expression for [tex] T [/tex] from the vertical balance of forces:
[tex] \dfrac{mg \sin(\theta)}{\cos(\theta)} = \dfrac{mv^2}{r} [/tex]
Simplify and solve for [tex] v [/tex]:
[tex] v^2 = rg \tan(\theta) [/tex]
[tex] v = \sqrt{rg \tan(\theta)} [/tex]
Now, substitute the values:
[tex] v = \sqrt{5.931 \times 9.8 \times \tan(34.4^\circ)} [/tex]
[tex] v = \sqrt{58.1238\times 0.6847142903} [/tex]
[tex] v = \sqrt{39.79819647} [/tex]
[tex] v = 6.308581177 [/tex]
[tex] v \approx 6.3 \, \textsf{m/s ( in 1 d.p.)} [/tex]
So, the speed of each seat is approximately [tex] 6.3 [/tex] m/s.
