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A circular-motion addict of mass80kgrides a Ferris wheel around in a vertical circle of radius10mat a constant speed of6.1m/s. (a) What is the period of the motion? What is the magnitude of the normal force on the addict from the seat when both go through (b) the highest point of the circular path and (c) the lowest point?

Short Answer

Expert verified
  1. The period of the motion is10s.

  2. The magnitude of the normal force on the addict from the seat when both go through the highest point of the circular path is4.9×102N

  3. The magnitude of the normal force on the addict from the seat when both go through the lowest point is1.1kN.

Step by step solution

01

Given

Mass=80kg,circleofradius=10m,speed=6.1m/s

02

Determining the concept

This problem is based on the concept of uniform circular motion. Uniform circular motion is a motion in which an object moves in a circular path with constant velocity. Also, it involves Newton’s second law of motion.


Formula:

The velocity in uniform circular motion is given by,

v=2πRT

where, v is the velocity, R is the radius and T is the time period

According to Newton’s second law of motion

FN-mg=mac

where,acis an acceleration, g is an acceleration due to gravity, m is mass,FN is the normal force and R is the radius.

03

(a) Determining the period of the motion

Using equation (i) the time can be written as,

T=2πR/V-2π(10M)/(6.1m/s)=10s

Hence, the period of the motion is 10 s.

04

(b)Determining the magnitude of the normal force on the addict from the seat when both go through the highest point of the circular path

In this case, Normal force FNis directed upward, gravitational force mg is to downward and acceleration is also directed to the down. Thus, by using Newton’s 2nd law,

FN-mg=mac(Centripetal acceleration is due to the circular motion)

FN=m(g-v2/R)=486N4.9×102N

Hence, the magnitude of the normal force on the addict from the seat when both go through the highest point of the circular path is4.9×102N.

05

(c) Determining the magnitude of the normal force on the addict from the seat when both go through the lowest point

Now, reverse both the normal force direction and the acceleration direction

Thus,

F=m(g+v2/R)-1081N1.1kN

Hence, the magnitude of the normal force on the addict from the seat when both go through the lowest point is1.1kN

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