ONONDAGA. 4. A woman rides on a Femis wheel of radius 16m that maintains the sam

Question

ONONDAGA. 4. A woman rides on a Femis wheel of radius 16m that maintains the same speed throughout its motion. To better understand physics, she takes along a digital bathroom scale (with memory) and sits on it. When she gets off the ride, she uploads the scale readings to a computer and creates a graph of scale reading versus time. Note that the graph has a minimum value of 510 N and a maximum value of 666 N. The acceleration due to gravity, g-9.80m/s (a At what moment did the woman pass through the lowest point of the circle? Explain. (b At what moment did the woman pass through the highest point of the circle? Explain. (co What is the woman's mass? (d What is the period of rotation of the wheel? 5. A child of 45 kg mass falls from a shed of 1.8-m high. During the fall his body is fully extended, but during the landing he bent his knees into a duck position. (a) At what velocity does he reach the ground? (b) If the landing lasts 0.45 s, what is the acceleration of the child during this sage of the motion? (e) Find the normal force exerted by the floor on the child during the landing. 6. Europa, a satellite of Jupiter, is believed to have an ocean of liquid water (with the possibility of life) beneath its icy surface. Europa is 3130 km in diameter and has a mass of weights ace of Europa. (b) If an astronaut 765 N on earth, what will be his weight on Europa? (c) What must be the velocity of an object, in order to remain stationary on a circular orbit of radius of 5250 km around Europa?

Explanation / Answer

4. a) at lowest point of circle,

Using Fnet = ma

N - mg = m a_c = m v^2 / r

N = mg + mv^2/r   ....This is the maximum possible reading thoughout the cirle.

so at reading 666N, she passes through lowest point.

b) At highest point,

mg - N = m v^2 / r

N = mg - m v^2/r ---- this lowest reading position

so 510 N position.

c) mg - mv^2/r = 510   .......(i)

m v^2/r + mg = 666 .....(ii)

(ii) + (i) =<

2mg = 666 + 510

m =60 kg

d) (ii) - (i) =>

2 m v^2 / r = 666 - 510

2 x 60 x v^2 / 16 = 156

v = 4.56 m/s

Related Questions

Navigate

• Browse (All)
• Browse O
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.