The least coefficient of static friction between the cat and the merry-go-round that will allow the cat to stay in place is approximately 0.15.
To find 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, we need to consider the forces acting on the cat. The static friction force between the cat and the merry-go-round provides the centripetal force required for circular motion.
The centripetal force is given by the equation F = mxω²xr, where m is the mass of the cat, ω is the angular velocity of the merry-go-round, and r is the radius.
By setting the static friction force equal to the centripetal force, we can solve for the coefficient of static friction. The equation is: μxN = mxω²xr, where μ is the coefficient of static friction and N is the normal force acting on the cat.
Since the cat is stationary, the normal force is equal to the gravitational force acting on the cat. Thus, N = mxg, where g is the acceleration due to gravity.
Substituting this into the equation, we get: μxmxg = mxω²xr. Solving for μ gives us: μ = (ω²xr)/g.
With the given values of ω = 1 revolution every 6 seconds (ω = 2π radians every 6 seconds) and r = 5.4 m, we can substitute into the equation and calculate μ. Using g = 9.8 m/s², we get: μ = (2π/6)²x5.4/9.8 ≈ 0.15.
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 0.15.