a small bead can slide without friction on a circular hoop that is in a vertical plane and has a radius of 0.100 m. The hoop rotates at a constant rate of 3.00 rev/s about a vertical diameter. Find the angle beta at which the bead is in vertical equilibrium. what would happen if the hoop rotates at 1.00 rev/s?
The bead is accelerating on the horizontal plane in a circle with radius r=RsinB. This is derived from the angle that the bead makes with the vertical. Once you have the radius of the particles trajectory you need to find the forces acting on it. The two forces acting on the bead are the force of gravity and the normal force the hoop exerts on the bead. The normal force is simply the force that is pointing towards the centre of the vertical hoop. Once you draw your FBD you can see now that the Force that's causing acceleration on the horizontal axis is simply NsinB. Using Newton's Second Law of motion, (F=ma) NsinB=ma. Acceleration of a moving body in a uniform circle is a=v^2/r. Therefore, by substitution, NsinB=mv^2/R. Now to find N we go back to our FBD. NcosB=mg, thus, N=mg/cosB. Subbing it back into the original equation we arrive at, gtanB=v^2/r. For v, v=2πr/T. Subbing that into the equation we arrive at, gtanB=(4π^2(RsinB)/T^2). r=RsinB. Rearrange, then find that cosB=(gT^2)/(4π^2)R). T=1/3s, So B= 74 degrees. To find what happens if the hoop rates at 1.00 rev/s we take a look at the equation, tanB=(4π^2R/gT^2)sinB. We notice that if the term inside the brackets is above one we can receive a numerical value for B other than 0. However, if T=1s, we cant find a value, therefore the only way to equate this is to have B=0. Therefore is bead stays stationary at the bottom.
Well, it sounds like this little bead is having a wild time on the circular hoop! Let's see if we can figure out its angle of equilibrium, shall we?
When the bead is in vertical equilibrium, it means that the gravitational force pulling it down is balanced by the normal force acting on it from the hoop. In other words, the net force in the vertical direction is zero.
The gravitational force on the bead is given by the formula Fg = mg, where m is the mass of the bead and g is the acceleration due to gravity. However, since the problem doesn't give us any information about the mass of the bead, let's just call it "mysterious mass" for now. Don't worry, this happens to be a favorite topic of physicists - they love mysteries!
Now, let's consider the normal force acting on the bead. This force is provided by the hoop and is directed perpendicular to the surface of the hoop. At the angle beta, which we're trying to find, the normal force has both horizontal and vertical components. The vertical component of the normal force has to exactly cancel out the gravitational force.
Since we're dealing with circular motion, we know that the acceleration of the bead is directed towards the center of the hoop. This acceleration is given by the formula a = rω², where r is the radius of the hoop and ω is the angular velocity.
The net force in the vertical direction is the difference between the vertical component of the normal force and the gravitational force. When the bead is in vertical equilibrium, this net force is zero. So we can set up the equation:
Vertical component of the normal force - gravitational force = 0
Here comes the fun part. Since we're in vertical equilibrium, the vertical component of the normal force is equal to the gravitational force. And since the bead is sliding without friction, it means that there are no other horizontal forces acting on it, so there is no horizontal component of the normal force.
So, we have:
mg - mg = 0
Surprise, surprise! The mysterious mass cancels out and we're left with zero. That means that the angle beta can be anything when the hoop rotates at a constant rate of 3.00 rev/s. The bead is as happy as a clam and can find equilibrium at any angle!
But wait, if the hoop rotates at 1.00 rev/s, well, things get interesting. The same equation still holds, but now we have a non-zero net force in the vertical direction. That means that the bead will no longer be in vertical equilibrium and it will experience a net force pulling it either up or down.
So, my friend, when the hoop rotates at 1.00 rev/s, the bead will be like a rebellious teenager, disobeying the laws of equilibrium and going wherever it pleases. It might slide up, it might slide down, but one thing is for certain - it won't be in vertical equilibrium!
I hope this helps you understand the situation, or at the very least, brings a smile to your face!
To find the angle β at which the bead is in vertical equilibrium, we can use the concept of centripetal force.
In vertical equilibrium, the net force on the bead must be zero. The gravitational force and the centripetal force are the only forces acting on the bead.
1. Gravitational force: The gravitational force acting on the bead is given by Fg = m * g, where m is the mass of the bead and g is the acceleration due to gravity.
2. Centripetal force: The centripetal force acting on the bead is given by Fc = m * ω^2 * r, where ω is the angular velocity of the hoop in radians per second and r is the radius of the hoop.
For the bead to be in vertical equilibrium, the centripetal force must equal the gravitational force.
Fc = Fg
m * ω^2 * r = m * g
The mass of the bead cancels out:
ω^2 * r = g
Substituting the given values:
ω = 3.00 rev/s (convert to radians per second: ω = 3.00 rev/s * 2π rad/rev ≈ 18.85 rad/s)
r = 0.100 m
18.85^2 * 0.100 = g
g ≈ 35.31 m/s^2
Next, we can find the angle β using trigonometry. The centripetal force is directed towards the center of the hoop and is balanced by the vertical component of the gravitational force.
cos(β) = Fg / Fc
cos(β) = g / (ω^2 * r)
cos(β) = 35.31 / (18.85^2 * 0.100)
cos(β) ≈ 0.198
β ≈ arccos(0.198)
β ≈ 78.8 degrees
So, the angle β at which the bead is in vertical equilibrium is approximately 78.8 degrees.
If the hoop rotates at 1.00 rev/s instead, the gravitational force would be greater than the centripetal force. This would cause the bead to slide down the hoop due to the unbalanced gravitational force. The bead would not be in vertical equilibrium and would move towards the bottom of the hoop.
To find the angle β at which the bead is in vertical equilibrium, we need to consider the forces acting on the bead.
When the bead is in vertical equilibrium, there are two forces acting on it: gravity (mg) pointing downward and the normal force (N) exerted by the hoop in the upward direction. The normal force provides the necessary centripetal force for the bead to move in a circular motion.
To start, let's consider the forces acting on the bead when the hoop rotates at a constant rate of 3.00 rev/s.
1. Centripetal Force: The centripetal force acting on the bead is provided by the normal force (N). It can be calculated using the formula:
N = m * ω² * r
Where m is the mass of the bead, ω is the angular velocity of the hoop (in rad/s), and r is the radius of the hoop.
In this case, since the bead is in vertical equilibrium, the centripetal force is equal to the weight of the bead:
N = mg
Combining both equations, we get:
mg = m * ω² * r
Rearranging the equation, we find:
ω² = g / r
2. Angular Velocity: The angular velocity of the hoop (ω) is related to the rotational speed (rev/s) by the formula:
ω = 2π * v
Where v is the rotational speed in rev/s. In this case, ω = 2π * 3.00 rad/s.
3. Angle β: The angle β at which the bead is in vertical equilibrium can be determined using trigonometry. We can use the relationship between the radius of the hoop and the angle β given by:
r * sin(β) = r
Simplifying, we find:
sin(β) = 1
Taking the inverse sine of both sides, we get:
β = arcsin(1) = 90°
Therefore, when the hoop rotates at a constant rate of 3.00 rev/s, the angle β at which the bead is in vertical equilibrium is 90 degrees.
Now, let's consider what would happen if the hoop rotates at 1.00 rev/s.
By applying the same steps as before, we can determine that the angular velocity (ω) of the hoop would be 2π * 1.00 rad/s.
Using the equation ω² = g / r, we find that the angular velocity would be smaller, resulting in a lower centripetal force. As a result, the bead would not be able to stay in vertical equilibrium, and it would slide down the hoop.
In conclusion, when the hoop rotates at 3.00 rev/s, the angle β at which the bead is in vertical equilibrium is 90 degrees. However, if the hoop rotates at 1.00 rev/s, the bead would not be in vertical equilibrium and would slide down the hoop.