Respuesta :
(a) 1.76 m/s^2
The centripetal acceleration of the child is given by:
[tex]a_c=\omega^2 r[/tex]
where
[tex]\omega[/tex] is the angular velocity
r is the radius of the wheel
The radius of the wheel is half the diameter:
[tex]r=\frac{d}{2}=\frac{20.0 m}{2}=10.0 m[/tex]
The wheel makes 4 revolution per minute, so the angular velocity is
[tex]\omega=4 rev/min[/tex]
Let's remind that
[tex]1 rev = 2 \pi rad[/tex]
[tex]1 min = 60 s[/tex]
So the angular velocity is
[tex]\omega=(4 rev/min) \cdot \frac{2 \pi rad/rev}{60 s/min}=0.42 rad/s[/tex]
So, the centripetal acceleration is
[tex]a_c=(0.42 rad/s)^2(10.0 m)=1.76 m/s^2[/tex]
(b) 509.1 N, upward
At the lowest point of the ride, we have the following forces:
- Normal force exerted by the seat on the child: N, upward
- Weight of the child: W = mg, downward
The resultant of these forces must be equal to the centripetal force, which is upward (towards the centre of the wheel), so we have the following equation
[tex]N-mg = ma_c[/tex]
From which we can find the normal reaction of the seat on the child:
[tex]N=m(g+a_c)=(44.0 kg)(9.81 m/s^2+1.76 m/s^2)=509.1 N[/tex]
(c) 354.2 N, upward
At the highest point of the ride, we have the following forces:
- Normal force exerted by the seat on the child: N, upward
- Weight of the child: W = mg, downward
The resultant of these forces must be equal to the centripetal force, which this time is downward (towards the centre of the wheel), so we have the following equation
[tex]mg-N = ma_c[/tex]
From which we can find the normal reaction of the seat on the child:
[tex]N=m(g-a_c)=(44.0 kg)(9.81 m/s^2-1.76 m/s^2)=354.2 N[/tex]
(d) 431.6 N, upward
When the child is halfway between the top and the bottom, the normal force exerted by the seat on the child is simply equal to the weight of the child; therefore we have:
[tex]N=mg=(44.0 kg)(9.81 m/s^2)=431.6 N[/tex]
Centripetal acceleration is towards the center. The force seat exerts on the child when the child is halfway between the top and bottom is 431.64 N.
What is centripetal acceleration?
The centripetal acceleration is caused due to change in direction of the body which is in a circular motion, the acceleration is towards the center of the circle. It is calculated using the formula,
[tex]a = \dfrac{v^2}{r}[/tex]
Given to us
Mass of the child, m = 44 kg
The angular velocity of the wheel, ω = 4 rev/ min. = 0.42 rev\sec
Diameter of the wheel, d = 20.0 m
The radius of the wheel, r = 10.0 m
A.) The centripetal acceleration of the child can be given as,
[tex]a = \dfrac{v^2}{r}[/tex]
Also, we know that the linear velocity is written as,
[tex]v = \omega \times r[/tex]
Substitute the value,
[tex]a = \dfrac{(\omega r)^2}{r} = \omega^2 r[/tex]
[tex]a = (0.42)^2 \times 10 = 1.764\ m/s^2[/tex]
B.) Force that the seat experts on the child,
At the point when the child is at the lowest point of the wheel,
there are three forces that will work on the child,
The normal force, that will act upwards on the child, N
The weight of the child that will act downwards, W = mg
The centripetal force that will act toward the center therefore upwards, [tex]F_c = m a[/tex]
Taking all the vertical forces,
[tex]\sum F_y = 0\\\\N + F_c = W\\\\N + ma = mg\\\\N = mg-ma\\\\N=m(g-a)\\\\\text{Substitute the values}\\\\N = 44(9.81-1.76)\\\\N = 354.2\ N[/tex]
C.) Force that the seat experts on the child,
At the point when the child is at the highest point of the wheel,
there are three forces that will work on the child,
The normal force, that will act upwards on the child, N
The weight of the child that will act downwards, W = mg
The centripetal force that will act toward the center therefore downwards, [tex]F_c = m a[/tex]
Taking all the vertical forces,
[tex]\sum F_y = 0\\\\N = F_c + W\\\\N = ma + mg\\\\N = mg+ma\\\\N=m(g+a)\\\\\text{Substitute the values}\\\\N = 44(9.81+1.76)\\\\N = 509.08\ N[/tex]
D.)C.) Force that the seat experts on the child,
At the point when the child is at the midway of the wheel,
there are three forces that will work on the child,
The normal force, that will act upwards on the child, N
The weight of the child that will act downwards, W = mg
The centripetal force that will act toward the center therefore Rightside, [tex]F_c = m a[/tex]
Taking all the vertical forces,
[tex]\sum F_y = 0\\\\N = W\\\\N = mg\\\\\text{Substitute the values}\\\\N = 44\times 9.81\\\\N = 431.64\ N[/tex]
Hence, the force seat exerts on the child when the child is halfway between the top and bottom is 431.64 N.
Learn more about Centripetal Force:
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