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Sagot :
Answer:
a) ω₁ = ω₂ = 3.7 rad/sec
b) Δθ₁ = Δθ₂ = 18.5 rad
c) d₁ = 14.5 m d₂ = 57.5 m
d) Fc1 = 273.9 N Fc2 = 1069.8 N
e) The boy near the outer edge.
Explanation:
a)
- Since the merry-go-round is a rigid body, any point on it rotates at the same angular speed.
- However, linear speeds of points at different distances from the center, are different.
- Applying the definition of angular velocity, and the definition of angle, we can write the following relationship between the angular and linear speeds:
[tex]v = \omega*r (1)[/tex]
- Since we know the value of v for the child near the outer edge, and the value of r for this point, we can find the value of the angular speed, as follows:
[tex]\omega = \frac{v_{out} }{r_{out} } = \frac{11.5m/s}{3.14m} = 3.7 rad/sec (2)[/tex]
- As we have already said, ωout = ωin = 3.7 rad/sec
b)
- Since the angular speed is the same for both childs, the angle rotated in the same time, will be the same for both also.
- Applying the definition of angular speed, as the rate of change of the angle rotated with respect to time, we can find the angle rotated (in radians) as follows:
- [tex]\Delta \theta = \omega * t = 3.7 rad/sec* 5.0 sec = 18.5 rad (3)[/tex]
⇒ Δθ₁ = Δθ₂ = 18.5 rad.
c)
- The linear distance traveled by each child, will be related with the linear speed of them.
- Knowing the value of the angular speed, and the distance from each boy to the center, we can apply (1) in order to get the linear speeds, as follows:
[tex]v_{inn} = \omega * r_{inn} = 3.7 rad/sec * 0.78 m = 2.9 m/s (4)[/tex]
vout is a given of the problem ⇒ vout = 11. 5 m/s
- Applying the definition of linear velocity, we can find the distance traveled by each child, as follows:
[tex]d_{inn} = v_{inn} * t = 2.9m/s* 5.0 s = 14.5 m (5)[/tex]
[tex]d_{out} = v_{out} * t = 11.5 m/s* 5.0 s = 57.5 m (6)[/tex]
d)
- The centripetal force experienced by each child is the force that keeps them on a circular movement, and can be written as follows:
[tex]F_{c} = m*\frac{v^{2}}{r} (7)[/tex]
- Replacing by the values of vin and rin, since m is a given, we can find Fcin (the force on the boy closer to the center) as follows:
[tex]F_{cin} = m*\frac{v_{in}^{2}}{r_{in}} = 25.4 kg* \frac{(2.9m/s)^{2} }{0.78m} = 273.9 N (8)[/tex]
- In the same way, we get Fcout (the force on the boy near the outer edge):
[tex]F_{cout} = m*\frac{v_{out}^{2}}{r_{out}} = 25.4 kg* \frac{(11.5m/s)^{2} }{3.14m} = 1069.8 N (9)[/tex]
e)
- The centripetal force that keeps the boys in a circular movement, is not a different type of force, and in this case, is given by the static friction force.
- The maximum friction force is given by the product of the coefficient of static friction times the normal force.
- Since the boys are not accelerated in the vertical direction, the normal force is equal and opposite to the force due to gravity, which is the weight.
- As both boys have the same mass, the normal force is also equal.
- This means that for both childs, the maximum possible static friction force, is the same, and given by the following expression:
- [tex]F_{frs} = \mu_{s} * m* g (10)[/tex]
- If this force is greater than the centripetal force, the boy will be able to hold on.
- So, as the centripetal force is greater for the boy close to the outer edge, he will have a more difficult time holding on.
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