Circular motion and friction problem?

Question

A cat dozes on a stationary merry-go-round, at a radius of 5.0 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 6.2 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 1

LOL -

Answer 2

If the cat stays in place without sliding, that means the cat is moving in a perfect circle of 5.0m radius, every 6.2 seconds.

That means you can determine the cat's acceleration. Use the formulas you've learned for calculating acceleration in circular motion. (Hint: one revolution is 2π radians; so the merry-go-round's angular speed is 2π radians per 6.2 seconds.)

Once you know the cat's acceleration, you can write a formula for the net force acting on the cat. Use Newton's 2nd law. (Use "m" for the cat's mass, which you don't know--but don't worry, the "m" will cancel out later).

The net force is entirely due to friction. Write a formula for the friction on the cat, in terms of the coefficient of static friction, μ. This should be a formula you've already learned.

Now set the two formulas for force equal to each other. That should give you an equation that you can solve for μ.

Answer 3

when the cat jumps on the merry-go-round, there is a centripetal acts on it. The direction of the centripetal is inward or perpendicular to the motion. Assume there is no friction, the cat will slide off the merry-go-round, parallel to the direction of the direction of the centripetal force. Because friction is always opposite of the motion, thus, the direction of friction is inward. Since the directions of centripetal and friction are both inward. Centripedal force IS friction force.

Fc = Friction

Fc = mv^2/r
friction = μ(Fnormal)

the normal force in this case is equal to the weight of the cat

Friction = μ(mg)

mv^2/r = μ(mg)

the mass cancel out
v^2/r = μg

μ = v^2 / (r g)

find the speed
V = circumference / Period

V = 2pi 5 / 6.2
V = 5.06 m/s

μ = (5.06)^2 / (5 x 9.8)
μ = .522

Answer 4

For the cat to move along a circle there must be centripetal acceleration.

The only acceleration is due to the friction between the cat and its seat and that is the centripetal acceleration.

Coefficient of static friction = centripetal acceleration /g = rω^2 /g

‘ω’ is the angular speed.

If T is the period. ω = 2 Л / T = 1.0134 rad/s

μ = 5 * (1.0134) ^2 / 9.8 = 0.524.

Answer 5

The normal force of the car must be equal to the centripetal force. Therefor, F(centripetal)=velocity squared, decided by the radius(40meters) then multiplied by the mass. All of that will then be set equal to> F(friction)=mgu