Given:
r=4.5 m
angular velocity
ω=2*π/7.2 rad/s
=0.8727 rad/s
When the cat is at position of 3 or 9 o'clock, the horizontal acceleration due to centripetal force is at its maximum.
The centripetal acceleration is
a=rω²
The centrifugal force is
F=ma=mrω²
where m=mass of cat
the normal reaction,
N=mg
g=acceleration due to gravity = 9.81 m/s²
So the minimum friction to keep the cat in place is
μ=F/N=rω²/g
Substitute numerical values and find μ.
A cat dozes on a stationary merry-go-round, at a radius of 4.5 m from the center of the ride. The operator turns on the ride and brings it up to its proper turning rate of one complete rotation every 7.2 s. What is the least coefficient of static friction between the cat and the merry-go-round that will allow the cat to stay in place, without sliding?
2 answers
butthole