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?
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.
butthole
To determine the minimum coefficient of static friction required for the cat to stay in place on the merry-go-round, we need to consider the forces acting on the cat.
First, let's analyze the forces at play. The two main forces acting on the cat are the gravitational force (mg) pulling the cat downwards and the static friction force (fs) exerted by the merry-go-round trying to prevent the cat from sliding.
The maximum static friction that can occur is given by the formula fs(max) = µs * N, where µs is the coefficient of static friction and N is the normal force. In this case, the normal force is equal to the gravitational force, N = mg.
For the cat to stay in place, the maximum static friction must be equal to or greater than the force trying to make the cat slide. The force trying to make the cat slide is the centripetal force (Fc), given by Fc = m * v^2 / r, where m is the mass, v is the velocity, and r is the radius.
In this case, we know the radius (r = 4.5 m) and the period of rotation (T = 7.2 s). The velocity (v) can be calculated using the equation v = 2πr / T, where π is the mathematical constant pi.
To determine the minimum coefficient of static friction (µs), we equate the centripetal force (Fc) to the maximum static friction force (fs(max)):
m * v^2 / r = µs * mg
Now we can plug in the values and solve for µs:
v = 2πr / T
v = 2 * π * 4.5 m / 7.2 s
v ≈ 3.14 m/s
Substituting the value of v into our equation:
m * (3.14 m/s)^2 / 4.5 m = µs * m * g
Simplifying and canceling m:
(3.14 m/s)^2 / 4.5 m = µs * g
Now we can solve for µs:
µs = (3.14 m/s)^2 / (4.5 m * g)
Finally, we plug in the value of g (acceleration due to gravity) which is approximately 9.8 m/s^2:
µs = (3.14 m/s)^2 / (4.5 m * 9.8 m/s^2)
Calculating this expression will give us the minimum coefficient of static friction required for the cat to stay in place on the merry-go-round.