Respuesta :
Answer:
a. 3.86 m/sĀ²
b. 332.64 N directed upwards towards the center of the ferris wheel.
c. 764.96 N directed downwards towards the center of the ferris wheel.
d. 589.84 N at 21.5Ā° counterclockwise from the horizontal direction.
Explanation:
a. Since the ferris wheel rotates 4 times per minute, its period, T = 60 s/4 = 15 s.
We now find it angular speed Ļ = 2Ļ/T = 2Ļ/15 = 0.418 rad/s
We then calculate its centripetal acceleration from a = rĻĀ² where r = radius of ferris wheel = 22.0 m.
So, a = 22 m Ć (0.418 rad/s)Ā² = 3.86 m/sĀ²
b. At the lowest point, the normal force, N and the centripetal force, F both act in opposite directions to the weight, mg of the object. So,
N + F = mg
N = mg - F Ā Ā Ā
N = mg - ma where a is the centripetal acceleration
N = m(g - a)
N = 56 kg(9.8 m/sĀ² - 3.86 m/sĀ²)
N = 56 kg Ć 5.94 m/sĀ²
N = 332.64 m/sĀ²
The normal force the seat exerts on the child is thus 332.64 N directed upwards towards the center of the ferris wheel.
c. At the highest point, the weight, mg of the object and the centripetal force, F both act in opposite directions to the normal force, N. So,
N = mg + F Ā Ā Ā
N = mg + ma where a is the centripetal acceleration
N = m(g + a)
N = 56 kg(9.8 m/sĀ² + 3.86 m/sĀ²)
N = 56 kg Ć 13.66 m/sĀ²
N = 764.96 N
The normal force the seat exerts on the child is thus 764.96 N directed downwards towards the center of the ferris wheel.
d. Half way between the top and the bottom of the ferris wheel, the normal force must balance the weight and the centripetal force so the child doesn't fall off. For this to happen, the normal force is thus the resultanf of the centripetal force and the weight of the child. Since these two forces are perpendicular at this instance,
N = ā[(mg)Ā² + (ma)Ā²] = mā(gĀ² + aĀ²) = 56 kgā[(9.8 m/sĀ²)Ā² + (3.86 m/sĀ²)Ā²] = 56 kgā[(97.196 (m/sĀ²)Ā² + 14.8996(m/sĀ²)Ā²] = 56kgā110.9396 (m/sĀ²)Ā² = 56 kg Ć 10.53 m/sĀ² = 589.84 N.
Since the centripetal force acts towards the center of the ferris wheel in the horizontal direction, it is equal to the horizontal component of the normal force, Also, the weight acts downwards and is equal to the vertical component of the normal force.
So, the direction of the normal force is gotten from
tanĪø = ma/mg = a/g
Īø = tanā»Ā¹(a/g) = tanā»Ā¹(3.86 m/sĀ² / -9.8 m/sĀ²) = tanā»Ā¹(-0.3939) = -21.5Ā°. Since the angle is it shows a counter clockwise direction.
So, the normal force is 589.84 N at 21.5Ā° counterclockwise from the horizontal direction.
(a) The centripetal acceleration of the child is 1.94 m/sĀ².
(b) The force the seat exert on the child at the lowest point of the ride is 657.4 N.
(c) The force the seat exert on the child at the highest point of the ride is 440.2 N.
(d) The force the seat exert on the child when the child is halfway between the top and bottom is 108.64 N.
The given parameters;
- mass of the child, m = 56 kg
- angular speed, Ļ = 4 rev/min
- diameter of the wheel, d = 22 m
- radius of the wheel, r = 11 m
The centripetal acceleration of the child is calculated as follows;
[tex]a_c = \omega ^2 r\\\\[/tex]
where;
Ļ Ā is angular speed of the wheel in rad/s
The angular speed of the wheel in rad/s is calculated as follows;
[tex]\omega = 4 \ \frac{rev}{\min} \times \frac{2\pi \ rad}{1 \ rev} \times \frac{1\min}{60 \ s} \\\\\omega = 0.419 \ rad/s[/tex]
The centripetal acceleration of the child is calculated as;
[tex]a_c = \omega^2 r\\\\a_c = (0.419)^2 \times 11\\\\a_c = 1.94 \ m/s^2[/tex]
The force the seat exert on the child at the lowest point of the ride is calculated as;
[tex]T_{bottom} = ma_c + mg\\\\T_{bottom} = m(a_c + g)\\\\T_{bottom} = 56(1.94 + 9.8)\\\\T_{bottom} = 657.4 \ N[/tex]
The force the seat exert on the child at the highest point of the ride is calculated as;
[tex]T_{top} = mg- ma_c\\\\T_{top} = m(g - a_c)\\\\T_{top} = 56(9.8 - 1.94)\\\\T_{top} = 440.2 \ N[/tex]
The force the seat exert on the child when the child is halfway between the top and bottom;
[tex]T = ma_c\\\\T = 56(1.94)\\\\T = 108.64 \ N[/tex]
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