A 3.6-kg cat is sitting on the edge a merry-go-round that is spinning at 2.7 rad/s. The diameter of the merry-go-round is 5.6 m. How large is the static friction between the cat’s paws and the surface of the merry-go-round if it does not slip off?
I'm going to assume you want the friction coefficient and not the force because the force is just:
F = r omega^2 = 5.6 * 2.7^2
mu = Normal /F = 3.6*9.8/F
To find the static friction between the cat's paws and the surface of the merry-go-round, we can use the concept of centripetal force.
The centripetal force is the force that keeps an object moving in a circular path. It is directed towards the center of the circle and is equal to the mass of the object multiplied by the square of its velocity divided by the radius of the circular path.
In this case, the centripetal force is provided by the static friction between the cat's paws and the surface of the merry-go-round, as it prevents the cat from slipping off.
The formula for centripetal force is given by:
F = (m * v^2) / r
Where:
F is the centripetal force,
m is the mass of the cat,
v is the velocity of the merry-go-round, and
r is the radius of the circular path.
First, let's calculate the radius of the circular path. The diameter of the merry-go-round is 5.6 m, so the radius would be half of that:
r = 5.6 m / 2 = 2.8 m
Next, we need to find the velocity of the merry-go-round. The problem states that the merry-go-round is spinning at a rate of 2.7 rad/s. The velocity of an object moving in a circle is equal to the radius of the circle multiplied by the angular velocity (in this case, the rate of rotation in rad/s). So:
v = r * ω
Where:
v is the velocity of the merry-go-round,
r is the radius of the circular path, and
ω is the angular velocity.
Substituting the values:
v = 2.8 m * 2.7 rad/s ≈ 7.56 m/s
Now we can substitute the known values into the formula for centripetal force:
F = (m * v^2) / r
F = (3.6 kg * (7.56 m/s)^2) / 2.8 m
Calculating this expression gives us:
F ≈ 28.91 N
So, the static friction between the cat's paws and the surface of the merry-go-round is approximately 28.91 Newtons.