a child sits on the edge of a spinning merry-go-round 3.2 m from its center. The coefficient of static friction between the child's bottom and the surface of the merry-go-round is .8. Find the maximum rate in revolutions per seconds, that the merry-go-round can be spun before the child slips off.
To find the maximum rate in revolutions per second at which the merry-go-round can be spun before the child slips off, we need to consider the forces acting on the child.
Let's start by analyzing the forces acting on the child. The two significant forces at play here are the normal force (N) and the force of static friction (f). Since the child is not slipping, the force of static friction is providing the centripetal force required to keep the child in circular motion.
The centripetal force (Fc) is given by the equation: Fc = m * a, where m is the mass of the child and a is the centripetal acceleration.
The centripetal acceleration is given by the equation: a = r * ω^2, where r is the radius of the merry-go-round and ω is the angular velocity (in radians per second).
The force of static friction (f) can be calculated using the equation: f = μ * N, where μ is the coefficient of static friction and N is the normal force.
Since the child is not moving vertically, the normal force (N) is equal to the child's weight (mg), where g is the acceleration due to gravity.
Now, let's put it all together:
The force of static friction (f) is equal to the centripetal force (Fc):
μ * N = m * a
Substituting the values for N and a:
μ * mg = m * r * ω^2
Mass (m) cancels out from both sides of the equation, giving us:
μ * g = r * ω^2
Solving for ω:
ω^2 = (μ * g) / r
ω = sqrt((μ * g) / r)
Finally, we can substitute the given values to find the maximum rate in revolutions per second:
ω = sqrt((0.8 * 9.8 m/s^2) / 3.2 m)
ω ≈ 3.13 rad/s
Since there are 2π radians in one revolution, we can convert the angular velocity to revolutions per second:
ω (in rev/s) = 3.13 rad/s / (2π rad/rev)
ω (in rev/s) ≈ 0.50 rev/s
Therefore, the maximum rate at which the merry-go-round can be spun before the child slips off is approximately 0.50 revolutions per second.