A cat dozes on a stationary merry-go-round, at a radius of 4.9 m from the center of the ride. Then the operator turns on the ride and brings it up to its proper turning rate of one complete rotation every 6.8 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?
To determine the least coefficient of static friction that will allow the cat to stay in place without sliding, we need to consider the forces acting on the cat.
In this scenario, there are two forces acting on the cat:
1. The centrifugal force: This force is experienced by the cat due to its rotation on the merry-go-round. It acts outward, away from the center of the rotation. The formula for the centrifugal force is Fc = m * ω^2 * r, where m is the mass of the cat, ω is the angular velocity in rad/s, and r is the radius of rotation.
2. The static friction force: This force is provided by the interaction between the cat and the merry-go-round. It acts inward, towards the center of the rotation, and opposes the centrifugal force. The formula for static friction is Fs = μs * N, where μs is the coefficient of static friction and N is the normal force acting on the cat.
To find the least coefficient of static friction, we need to set up an equation that balances the centrifugal force and the static friction force:
Fc = Fs
Substituting the respective formulas, we have:
m * ω^2 * r = μs * N
Now, we need to find the value of N. The normal force is equal to the weight of the cat, which is given by:
N = m * g,
where g is the acceleration due to gravity.
Substituting N in the equation, we get:
m * ω^2 * r = μs * m * g
The mass (m) cancels out:
ω^2 * r = μs * g
Finally, we can solve for the coefficient of static friction, μs:
μs = (ω^2 * r) / g
Now, let's substitute the given values:
ω = 2π / T (T is the period of rotation, which is 6.8 s)
r = 4.9 m
g = 9.8 m/s^2
Calculating:
ω = 2π / 6.8 ≈ 0.92 rad/s
μs = (0.92^2 * 4.9) / 9.8 ≈ 0.31
Therefore, 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 is approximately 0.31.