A single bead can slide with negligible friction on a wire that is bent into a circular loop of radius 15.0 cm, as in Figure P6.68. The circle is always in a vertical plane and rotates steadily about its vertical diameter with (a) a

period of 0.450 s. The position of the bead is described by the angle θ that the radial line, from the center of the loop to the bead, makes with the vertical. At what angle up from the bottom of the circle can the bead stay motionless relative to the turning circle? (b) What If? Repeat the problem if the period of the circle’s rotation is 0.850 s.

To solve this problem, we need to analyze the forces acting on the bead and determine the conditions for it to stay motionless relative to the turning circle.

(a) For a bead to stay motionless relative to the turning circle, the net force acting on the bead must be zero. This means that the gravitational force pulling the bead downward must be balanced by the net horizontal component of the force exerted by the wire on the bead.

To determine the angle θ at which the bead can stay motionless, we need to calculate the net horizontal component of the force exerted by the wire on the bead. This force is provided by the tension in the wire.

Using the concept of centripetal force, we can find the tension in the wire as follows:

The centripetal force is given by:

F_c = m * a_c

where:
m is the mass of the bead
a_c is the centripetal acceleration

The centripetal acceleration can be calculated from the period of rotation as follows:

a_c = (2 * π * R) / T^2

where:
R is the radius of the circular loop
T is the period of rotation

Now, let's calculate the centripetal acceleration:

a_c = (2 * π * 0.15) / (0.450)^2
= 11.02 m/s^2

The centripetal force is the horizontal component of the tension. Since the wire forms an angle θ with the vertical, the horizontal component can be written as:

F_h = T * cos(θ)

Setting the net force equal to zero:

F_h - mg = 0

Substituting F_h and solving for θ:

T * cos(θ) - mg = 0

T * cos(θ) = mg

θ = arccos(mg / T)

Now, let's calculate θ by substituting the given values:

θ = arccos((m * g) / T)

(b) Let's repeat the same calculations for a period of 0.850 s. We will use the same formula as above, but substitute the new T value:

θ = arccos((m * g) / T)

θ = arccos((m * g) / 0.850)

To find the angle at which the bead stays motionless, we need to consider the forces acting on the bead. Since the circle rotates steadily with a constant period, the bead experiences two forces: the gravitational force (mg) and the centripetal force (mv^2/r), where m is the mass of the bead, g is the acceleration due to gravity, v is the velocity of the bead, and r is the radius of the circle.

(a) Let's consider the first question where the period of rotation is 0.450 s. At the angle where the bead stays motionless, the gravitational force and the centripetal force must balance each other out. At this angle, the gravitational force will act vertically downwards, while the centripetal force will act radially towards the center of the circle.

Since the bead is motionless, the net force acting on it must be zero. Therefore, we can equate the magnitude of the two forces:

mg = mv^2/r

The mass of the bead cancels out, so we are left with:

g = v^2/r

We know that the period (T) is related to the velocity (v) and the radius (r) through the equation v = 2πr/T. By substituting this into the equation, we get:

g = (2πr/T)^2 / r

Simplifying further, we get:

g = 4π²r / T²

Now, we can solve for the angle θ using the equation sin(θ) = opposite/hypotenuse, where the opposite side is the vertical component of the gravitational force (mg) and the hypotenuse is the total gravitational force (mg):

sin(θ) = mg_vertical / mg_total

sin(θ) = (mg * sin(θ)) / (mg)

sin(θ) = sin(θ)

This means that the angle θ is the same as the angle of the vertical line, which is 90 degrees or π/2 radians.

Therefore, the bead stays motionless at an angle 90 degrees up from the bottom of the circle.

(b) Now let's consider the second question where the period of rotation is 0.850 s. We can repeat the same steps to find the angle θ.

Using the equation we derived earlier, g = 4π²r / T², we can substitute the new period of rotation:

g = 4π²r / (0.850 s)²

Simplifying further, we get:

g = 4π²r / 0.7225 s²

Using the same equation sin(θ) = mg_vertical / mg_total, we can find the angle θ by calculating sin⁻¹(g_vertical / g_total):

θ = sin⁻¹(g_vertical / g_total)

Note: To calculate the gravitational force mg, we need to know the mass of the bead. Since the mass is not given in the problem statement, we are unable to provide an exact numerical answer. However, you can substitute the appropriate value of mg into the equation to find the angle in radians.

I hope this explanation helps!

(a) To find the angle at which the bead can stay motionless relative to the turning circle, we need to consider the forces acting on the bead.

At the bottom of the circle, the bead experiences a gravitational force pulling it downwards. As it moves up, the gravitational force decreases and at a certain angle, it balances out with the normal force exerted by the wire.

Using some humor, let's call this angle the "sweet spot" where the bead can chill and not move.

Now, let's get serious and calculate this angle.

At the bottom, the gravitational force is given by the weight of the bead, which is m*g, where m is the mass of the bead and g is the acceleration due to gravity.

At the sweet spot, the normal force cancels out the gravitational force, so we have:

m*g = N

where N is the normal force.

Since the bead is motionless, the net force acting on it in the radial direction must be zero.

At the sweet spot, the only forces acting on the bead in the radial direction are the normal force and the component of the gravitational force.

The component of the gravitational force in the radial direction is given by:

mg * cos(θ)

Therefore, the equation becomes:

N - mg * cos(θ) = 0

Rearranging the equation, we get:

N = mg * cos(θ)

Substituting this into the previous equation, we have:

mg * cos(θ) - mg = 0

Simplifying, we find:

cos(θ) = 1

Since the cosine of any angle is 1 only when the angle is 0, we conclude that the bead can stay motionless relative to the turning circle at the bottom, where θ = 0 degrees.

(b) Let's repeat the problem with a period of 0.850 s.

Using the same approach as before, we find that the bead can stay motionless relative to the turning circle at the bottom, where θ = 0 degrees. The period of rotation does not affect this result.

So, no matter how long the period of rotation is, the bead can always stay chill at the bottom of the circle without moving.

Just give me the damn answer