A 40-kg child rides on a Ferris Wheel of radius 20 m. The time for the Ferris Wh

Question

A 40-kg child rides on a Ferris Wheel of radius 20 m. The time for the Ferris Wheel to make a full rotation is 30 sec. (a) What are the forces acting on the child at the top of the ride? Show these forces in the figure to the right. (b) Which force is the greatest? What is the direction of the net force? (c) What is the magnitude and direction of the acceleration of the child at the top? (d) Apply Newton's 2 law to determine the normal force exerted by the seat on the child at the top. (e) The normal force is the 'effective' weight of the child. How does this compare with the child's actual weight?

Explanation / Answer

Velocity of child

V= 2*3.14* 20/ 30= 4.187 m/s

a)

There are two forces that are acting on the child, gravitational force and centripital force.

Fg= mg= 40*9.8= 392 N

Fc = mv^2/r= 40*(4.187^2)/20= 35.06 N

========

b)

Gravitational force is greater than centripital force

Net force directed to the down direction.

=======

c)

a= F/m= (392-35.06)/40= 8.92 m/s^2

Directed downwards

=====

d)

Normal force,

N= 8.92*40= 356.94 N

======

e) F= 356.94/392= 0.91 * weight

======

Comment in case any doubt.. good luck

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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