A cat sits on a stationary merry go round, at a radius of 5.4 m from the center of the ride. Then the operator turns on the ride and brings it up to its proper turning rate of one complete rotation every 6.0 s. What is 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?

Answer

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 can follow these steps:

### Step 1: Determine the angular speed (ω)
The merry-go-round completes one full rotation every 6 seconds. The angular speed can be calculated using the formula:
\[ \omega = \frac{2\pi}{T} \]
where \(T\) is the time period for one complete rotation.

Substituting \(T = 6 \, \text{s}\):
\[ \omega = \frac{2\pi}{6} = \frac{\pi}{3} \, \text{rad/s} \approx 1.05 \, \text{rad/s} \]

### Step 2: Calculate the centrifugal force (Fc)
The centrifugal force acting on the cat can be expressed as:
\[ F_c = m \omega^2 r \]
where \(m\) is the mass of the cat, \(r\) is the radius (5.4 m), and \(\omega\) is the angular speed calculated in Step 1.

### Step 3: Set up the static friction force (Fs)
The maximum static friction force that can act on the cat is given by:
\[ F_s = \mu_s N \]
where \(\mu_s\) is the coefficient of static friction and \(N\) is the normal force. Since there is no vertical motion, the normal force \(N\) is equal to the weight of the cat:
\[ N = mg \]

Thus, we can write:
\[ F_s = \mu_s mg \]

### Step 4: Equate the forces
For the cat to stay in place without sliding, the maximum static friction must be equal to or greater than the centrifugal force:
\[ F_s \geq F_c \]
Substituting the expressions for \(F_s\) and \(F_c\):
\[ \mu_s mg \geq m \omega^2 r \]

### Step 5: Simplify the equation
We can cancel \(m\) from both sides (assuming \(m \neq 0\)):
\[ \mu_s g \geq \omega^2 r \]

### Step 6: Solve for the coefficient of static friction (μs)
Rearranging gives:
\[ \mu_s \geq \frac{\omega^2 r}{g} \]

Substituting the values:
- \(\omega \approx 1.05 \, \text{rad/s}\)
- \(r = 5.4 \, \text{m}\)
- \(g = 9.8 \, \text{m/s}^2\)

Calculating:
\[ \mu_s \geq \frac{(1.05)^2 \times 5.4}{9.8} \]
Calculating \(1.05^2 \approx 1.1025\):
\[ \mu_s \geq \frac{1.1025 \times 5.4}{9.8} \approx \frac{5.95}{9.8} \approx 0.607 \]

Thus, the least coefficient of static friction is approximately:
\[ \mu_s \approx 0.61 \]

### Final Answer:
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 **0.61**.

---