A small button placed on a horizontal rotating platform with diameter 0.320 m will revolve with the platform when it is brought up to a speed of 40.0 rev min, provided the button is no more than 0.150 m from the axis. (a) What is the coefficient of static friction between the button and the platform

0.2687 approximately 0.27

Explanation:

Diameter = 0.320

Speed = 40.0 rev/min

We are required to find coefficient of static friction between friction and button

The radius can be calculated as

0.320/2

= 0.160m

Then we have the rotational speed w = 40rev/min x 2pi/60

= 4.19 rad/s

umg = mrw²

u = mrw²/mg

u = rw²/g (1)

g = 9.8

When we put values into equation 1

0.150m x 4.19² / 9.8

= 0.150m x 17.5561 /9.8

= 0.2689

This is approximately 0.27

d₁ = displacement of tern from arctic to antarctic = 11000 miles

d₂ = displacement of tern from antarctic to arctic =   - 11000 miles

d = total displacement for the journey

total displacement for the journey is given as

d = d₁ + d₂

d = 11000 + (-11000)

d = 0 miles

t₁ = time of travel from arctic to antarctic = 3 months = 3 x 30 days = 90 days

t₂ = time of travel from antarctic to arctic = 3 months = 3 x 30 days = 90 days

t = total time of travel

total time of travel is given as

t = t₁ + t₂

t = 90 + 90

t = 180 days

average velocity is given as

average velocity = total displacement/ total time

v = d/t

v = 0/180

v = 0 miles/day