Question

a cat is riding a merry-go-round that has a radius of 6 m and completes one rotation every 5 seconds. what is the minimum static coefficient of friction necessary between the cat and ride to keep it in place without sliding off?

Answer 1

To find the minimum static coefficient of friction required to keep a cat on a rotating merry-go-round, the steps involve computing the merry-go-round's angular velocity, the centripetal acceleration, and equating the centripetal force to the maximum static frictional force to solve for the coefficient of friction.

Step-by-step explanation:

The student's question pertains to the concept of centripetal force and the necessary coefficient of static friction to prevent a cat from sliding off a merry-go-round. The merry-go-round in question has a radius of 6 meters and completes one rotation every 5 seconds. To determine the minimum static coefficient of friction, we must consider the centripetal acceleration that the cat experiences while the merry-go-round is in motion.

1. First, calculate the angular velocity of the merry-go-round, using the formula ω = 2π/T, where T is the period of rotation (5 seconds).
2. Next, use the centripetal acceleration formula, a_c = ω² * r, where r is the radius (6 meters).
3. Then, apply the equation for centripetal force, F_c = m * a_c, where m is the mass of the cat. In this problem, the mass of the cat is not provided, but for static friction calculation, it is sufficient to equate this force with the maximum force of static friction, which is F_friction = μ * m * g, where μ is the coefficient of static friction and g is the acceleration due to gravity (9.81 m/s²).
4. Set F_c equal to F_friction to solve for the coefficient of static friction: μ = F_c / (m * g).
5. Since the mass of the cat cancels out in the equation, we can find the coefficient of static friction without knowing the cat's mass. Calculate μ using the values obtained for centripetal acceleration and gravity.
   

The minimum static coefficient of friction is determined by the centripetal acceleration that needs to be provided by the frictional force to keep the cat from sliding off. By using the above steps, one would be able to calculate the exact value necessary to prevent the cat from slipping off the merry-go-round.