Can you show all your work with formulas used? I also put in text form if that n

Question

Can you show all your work with formulas used? I also put in text form if that new picture still isn't clear enough for you.

At the bottom of the ride, what is the parallel component of the rate of change of the rider's momentum?

At the bottom of the ride, what is the perpendicular component of the rate of change of the rider's momentum?

At the bottom of the ride, what is the vector gravitational force exerted by the Earth on the rider?

At the bottom of the ride, what is the vector force exerted by the seat on the rider?

Next consider the situation at the top of the ride. At the top of the ride, what is the parallel component of the rate of change of the rider's momentum?

At the top of the ride, what is the perpendicular component of the rate of change of the rider's momentum?

At the top of the ride, what is the vector gravitational force exerted by the Earth on the rider?

At the top of the ride, what is the vector force exerted by the seat on the rider?

A rider feels heavier if the electric, interatomic contact force of the seat on the rider is larger than the rider's weight mg (and the rider sinks more deeply into the seat cushion). A rider feels lighter if the contact force of the seat is smaller than the rider's weight (and the rider does not sink as far into the seat cushion).

Does a rider feel heavier or lighter at the bottom of a Ferris wheel ride?

heavierlighter    

Does a rider feel heavier or lighter at the top of a Ferris wheel ride?

heavierlighter   

d|p| dt p = kg · m/s/s ertkcs, t the battom of the ide, what is the parallel component of the re nchang t rider's moentum? At the bottom or the nde.hM IS the perpen tadar oomponent ore rat or cnarge of the rders momentum? -I -... At the tottom ef tha ride, what is the vector taonal force setad b the Earth on the rider? At the boom he nde, wha isthe cor anse eeed by the sonerider? Nest coaidor the situation tha top ot the rida At the top of the nda, hat is the paralcl component af the rate of changa of the rider's momentu At the top ef the nde, hx is te pependoular corpanent o the rate of chargeof he dt's memum? t the top of the nde, what is t ecto grwuitabanal force certed by th Earth on the rider Mt the top of the ride, whst is the vector forte exerted by the seat on the ricder? A rider es heaver r the e ectric, in rotor ic co tot force o the sest on the rlder is larger than e de-'s weight g ond the ider sinks morn oephr into the seat c cn ,A der es gro rif the contact once of the seattmaler th the nder's we ht srd t e rider oes not sn as ar res the seat cushion Does nder ,eel heaver or Ighter at the bettom of Femi5-nee1mce? Coa a rideier o lihtar at the top of a Fom whaal ride? lighter

Explanation / Answer

since the ferris wheel is rotating at constant speed , hence parallel component of rate of change on momentum or the force will be 0

for perpendicular component :

w = angular speed = 2pi/T = 2 x 3.14 /10 = 0.628 rad/s

r = radius = 9 m

m = mass = 56 kg

Fn = perpendicular component

force equation at the bottom is given as

Fn - mg = m r w2

Fn = mg + m r w2 = (56) (9.8 + 9 (0.628)2) = 747.6 N

Fgrav = - mg = - (56) (9.8) = - 548.8 N

Fseat = Fn = mg + m r w2 = (56) (9.8 + 9 (0.628)2) = 747.6 N

at the top : parallel component = 0

at the top : perpendicular component

force equation at the top is given as

mg - Fn = m r w2

Fn = mg - m r w2 = (56) (9.8 - 9 (0.628)2) = 350.03 N

Fgrav at Top = - mg = - (56) (9.8) = - 548.8 N

Fseat at Top = Fn = mg - m r w2 = (56) (9.8 - 9 (0.628)2) = 350.03 N

Related Questions

Navigate

• Browse (All)
• Browse C
• Subjects

---

---

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