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Chapter 6: Q96P (page 148)

A child places a picnic basket on the outer rim of a merry-go-round that has a radius of 4.6 mand revolves once every 30s. (a) What is the speed of a point on that rim? (b)What is the lowest value of the coefficient of static friction between basket and merry-go-round that allows the basket to stay on the ride?

Short Answer

Expert verified

a) Speed of a point on rim is 0.96m/s

b)渭s=0.021

Step by step solution

01

Given

Merry-go-round that has a radius of 4.6 m and revolves once every 30 s

02

Understanding the concept

The problem is based on the centrifugal force. Centrifugal force is the apparent outward force on a mass when it is rotated. It also deals with the static friction.

Formula:

Centrifugal force is given by

F=mv2R

and the maximum value of static friction is given by

fs,max=渭smg.

03

Calculate the speed of a point on that rim

(a)

The distance traveled in one revolution is

2蟺搁=2蟺4.6m=29m.

The speed is

v=29m/30s=0.96m/s

04

Calculate the lowest value of the coefficient of static friction between basket and merry-go-round that allows the basket to stay on the ride 

(b)

According to Newton鈥檚 second law

fs=mv2/R=m0.20

As N = mg in this situation, the maximum possible static friction is,

fs,max=渭smg.

Equating this withlocalid="1660973301753" fs=m0.20渭s=0.20/9.8=0.021

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