The diagram represents wavelengths of different colors of visible light that pass by in a specific time period.

suppose the ball has the smallest possible frequency that allows it to go all the way around the circle. what tension in the str

tatyana61 [14]

The complete question is missing, so i have attached the complete question.

Answer:

A) FBD is attached.

B) The condition that must be satisfied is for ω_min = √(g/r)

C) The tension in the string would be zero. This is because at the smallest frequency, the only radially inward force at that point is the weight(force of gravity).

Explanation:

A) I've attached the image of the free body diagram.

B) The formula for the net force is given as;

F_net = mv²/r

We know that angular velocity;ω = v/r

Thus;

F_net = mω²r

Now, the minimum downward force is the weight and so;

mg = m(ω_min)²r

m will cancel out to give;

g = (ω_min)²r

(ω_min)² = g/r

ω_min = √(g/r)

The condition that must be satisfied is for ω_min = √(g/r)

C) The tension in the string would be zero. This is because at the smallest frequency, the only radially inward force at that point is the weight(force of gravity).

5 0

2 years ago

An amusement park ride called the Rotor debuted in 1955 in Germany. Passengers stand in the cylindrical drum of the Rotor as it

steposvetlana [31]

Answer:

μs = 0.36

Explanation:

While the drum is rotating, the riders, in order to keep in a circular movement, are accelerated towards the center of the drum.
This acceleration is produced by the centripetal force.
Now, this force is not a different type of force, is the net force acting on the riders in this direction.
Since the riders have their backs against the wall, and the normal force between the riders and the wall is perpendicular to the wall and aiming out of it, it is easily seen that this normal force is the same centripetal force.
In the vertical direction, we have two forces acting on the riders: the force of gravity (which we call weight) downward, and the friction force, that will oppose to the relative movement between the riders and the wall, going upward.
When this force be equal to the weight, it will have the maximum possible value, which can be written as follows:

$F_{frmax} = \mu_{s}* F_{n} = m * g (1)$