A child sits on a rotating merry-go-round, 2.19 m from its center. If the speed of the child is 2.32 m/s, what is the minimum coefficient of static friction between the child and the merry-go-round that will prevent the child from slipping?
ΣFy=0
Normal Force = mass*gravity
N=m*g
ΣFx=ma
Force of Friction = mass*velocity^2/radius
F=m*v^2/r
Force of Friction = Coefficient of static friction*Normal force
F=Us*N
So...
Us*m*g=m*v^2/r
Us=v^2/(r*g)
Well, if the child starts slipping, it's going to be a real "merry-go-round" of trouble! But don't worry, I'm here to help with some humor-infused physics.
To figure this out, we can start with the centripetal force required to keep the child moving in a circle. The formula is Fc = m * v^2 / r, where Fc is the centripetal force, m is the mass of the child, v is the speed, and r is the radius.
Since we want to find the minimum coefficient of static friction, we know that static friction provides the necessary centripetal force. So we can equate the maximum static friction force (μs * m * g) to the centripetal force.
μs * m * g = m * v^2 / r
Here, μs is the coefficient of static friction, m is the mass of the child, v is the speed, r is the radius, and g is the acceleration due to gravity.
Now we can solve for μs:
μs = v^2 / (r * g)
Plug in the given values: v = 2.32 m/s and r = 2.19 m, and use the acceleration due to gravity of around 9.8 m/s^2.
μs = (2.32 m/s)^2 / (2.19 m * 9.8 m/s^2)
After some calculations, we find that the minimum coefficient of static friction is approximately 0.111. So, to prevent the child from slipping, the minimum coefficient of static friction between the child and the merry-go-round should be greater than 0.111.
Now let's hope that the child holds on tight and doesn't become a "merry-go-round" projectile!
To determine the minimum coefficient of static friction between the child and the merry-go-round, we need to analyze the forces acting on the child.
1. Centripetal Force (Fc):
The centripetal force acting on the child is provided by the static friction force between the child and the merry-go-round. This force is directed towards the center of the merry-go-round, keeping the child in circular motion.
2. Weight Force (mg):
The weight force of the child acts vertically downward, and its magnitude is given by the formula Fw = mg, where m is the mass of the child and g is the acceleration due to gravity.
3. Static Friction Force (Fs):
The static friction force acts horizontally inward towards the center of the merry-go-round, opposing the child's tendency to slip tangentially. The maximum value of static friction force can be given by the formula Fs(max) = μs * Fn, where μs is the coefficient of static friction and Fn is the normal force.
Now, to find the minimum coefficient of static friction (μs), we can set up the following equation:
Fs(max) = Fc
μs * Fn = m * (v^2 / r)
Since the child is not slipping, the static friction force must attain its maximum value:
Fs(max) = Fs = μs * Fn
Also, the normal force (Fn) acting on the child is equal to its weight (mg):
Fn = mg
By substituting these equations, we get:
μs * mg = m * (v^2 / r)
Now, we can rearrange the equation to isolate the coefficient of static friction (μs):
μs = (m * (v^2 / r)) / mg
Cancelling the mass (m) terms, the equation simplifies to:
μs = (v^2 / (r * g))
Substituting the given values into the equation:
μs = (2.32^2 / (2.19 * 9.8))
Calculating this expression will give you the minimum coefficient of static friction required to prevent the child from slipping.