A cat dozes on a stationary merry-go-round, at a radius of 4.9 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 6.8 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?
Well, first of all, I must say that it's quite impressive that the cat decided to take a nap on a merry-go-round. I guess they really know how to find the perfect spot for a good snooze!
Now, let's tackle the physics of the situation. To determine the least coefficient of static friction that will keep the cat in place, we need to consider the forces acting on the cat.
When the merry-go-round starts rotating, the cat experiences a centripetal force pulling it towards the center. This force is provided by the friction between the cat's paws and the surface of the merry-go-round.
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 cat (which is just the circumference of the circle divided by the time taken for one complete rotation)
- r is the radius of the merry-go-round
Now, if the cat is not sliding, the frictional force should be equal to the centripetal force. The frictional force can be calculated using:
F_friction = μ * N
Where:
- F_friction is the frictional force
- μ is the coefficient of static friction
- N is the normal force (equal to the weight of the cat)
So, we can set up a relationship between the centripetal force and the frictional force:
m * (v^2 / r) = μ * N
Now, since the cat is not sliding, N = mg, where g is the acceleration due to gravity.
Plugging in the values and solving for μ, we get:
μ = (m * v^2) / (r * g)
With the values given, we can calculate μ. Let's hope the answer doesn't make the cat slip into an existential crisis!
Calculating... calculating...
*drum roll*
The minimum coefficient of static friction needed for the cat to stay in place is approximately 0.363.
So, in other words, as long as the coefficient of static friction between the cat and the merry-go-round is greater than 0.363, the cat will stay put. Otherwise, the cat might go for a wild ride, sliding off the merry-go-round like a cartoon character!
To find the least coefficient of static friction that will prevent the cat from sliding on the merry-go-round, we need to consider the forces acting on the cat.
The only force that can cause the cat to slide is the frictional force between the cat and the merry-go-round. The maximum static friction force can be calculated using the formula:
\(f_s = \mu_s \cdot N\)
Where \(f_s\) is the static friction force, \(\mu_s\) is the coefficient of static friction, and \(N\) is the normal force.
In this case, the normal force acting on the cat is equal to its weight. The weight of the cat can be calculated using the formula:
\(W = m \cdot g\)
Where \(m\) is the mass of the cat and \(g\) is the acceleration due to gravity.
We can find the mass of the cat by dividing its weight by the acceleration due to gravity:
\(m = \frac{W}{g}\)
The acceleration due to gravity is approximately \(9.8 \, \text{m/s}^2\).
Now, let's calculate the mass of the cat:
\(m = \frac{W}{g} = \frac{m \cdot g}{g} = m\)
We can now substitute the mass of the cat into the formula for the maximum static friction force:
\(f_s = \mu_s \cdot N = \mu_s \cdot m \cdot g\)
The maximum static friction force must provide the necessary centripetal force to keep the cat on the merry-go-round. The centripetal force can be calculated using the formula:
\(f_c = m \cdot \omega^2 \cdot r\)
Where \(f_c\) is the centripetal force, \(m\) is the mass of the cat, \(\omega\) is the angular velocity of the merry-go-round, and \(r\) is the radius.
Since the cat is stationary on the merry-go-round, the maximum static friction force must equal the centripetal force:
\(f_s = f_c\)
Now we can substitute the expressions for the maximum static friction force and the centripetal force:
\(\mu_s \cdot m \cdot g = m \cdot \omega^2 \cdot r\)
Simplifying this equation, we find:
\(\mu_s = \frac{\omega^2 \cdot r}{g}\)
Now we can substitute the given values into the equation to find the coefficient of static friction:
\(\mu_s = \frac{\left(\frac{2\pi}{T}\right)^2 \cdot r}{g}\)
Where \(T\) is the time for one complete rotation of the merry-go-round.
Given:
Radius, \(r = 4.9 \, \text{m}\)
Time for one complete rotation, \(T = 6.8 \, \text{s}\)
Acceleration due to gravity, \(g = 9.8 \, \text{m/s}^2\)
Substituting these values into the equation:
\(\mu_s = \frac{\left(\frac{2\pi}{6.8}\right)^2 \cdot 4.9}{9.8}\)
\(\mu_s = \frac{\left(\frac{2\pi}{6.8}\right)^2 \cdot 4.9}{9.8}\)
Evaluating this expression:
\(\mu_s \approx 0.388\)
Therefore, 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 0.388.
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 need to consider the forces acting on the cat.
First, let's analyze the forces on the cat in the rotating frame of reference. In this frame of reference, there are two forces acting on the cat:
1. The gravitational force pulling the cat downward, which can be represented by its weight (mg), where m is the mass of the cat and g is the acceleration due to gravity (approximately 9.8 m/s^2).
2. The centrifugal force pushing the cat outward, which is experienced due to the rotation of the merry-go-round. The centrifugal force can be calculated using the formula F = m * ω^2 * r, where m is the mass of the cat, ω is the angular velocity (2π/T, where T is the period of rotation), and r is the radius of rotation.
In this problem, we are looking for the least coefficient of static friction, which means we need to find the condition where the frictional force is at its minimum.
The frictional force acting on the cat is equal to the product of the coefficient of static friction (μs) and the normal force (N), where N is the force exerted by the merry-go-round on the cat and is equal to the sum of the gravitational force and the centrifugal force.
Now, let's calculate the normal force (N):
N = mg + m * ω^2 * r
Substituting the given values into the formula:
N = m * (g + ω^2 * r)
To find the minimum coefficient of static friction (μs_min), we need to find the condition where the frictional force is equal to the centrifugal force. This is because any value of the coefficient of static friction less than this will result in the cat sliding.
Setting the frictional force equal to the centrifugal force:
μs * N = m * ω^2 * r
Substituting N with its value:
μs * (m * (g + ω^2 * r)) = m * ω^2 * r
Now, solving for μs_min:
μs_min = (m * ω^2 * r) / (m * (g + ω^2 * r))
Canceling out the mass (m) and rearranging the equation:
μs_min = ω^2 * r / (g + ω^2 * r)
Substituting the given values:
ω = 2π / T,
r = 4.9 m, and
T = 6.8 s
ω = 2π / 6.8 = 0.9202 rad/s
μs_min = (0.9202^2 * 4.9) / (9.8 + 0.9202^2 * 4.9)
Calculating the value:
μs_min ≈ 0.268
Therefore, 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 0.268.