A small sphere has a hole drilled in it and it is placed onto a wire. The wire is bent into a circle of radius 15 centimeters. The sphere can slide freely without friction around the circular wire. The circular wire is held vertically.
(a) If the circular loop of wire is spun around its vertical diameter with a period of 0.45 seconds, at what angle up from the bottom will the sphere slide so that it stays at a steady height?
(b) Repeat the problem using a period of 0.85 seconds.
To find the angle at which the sphere will slide and maintain a steady height, we need to equate the gravitational force with the centrifugal force acting on the sphere.
(a) Let's start with the first case where the period is 0.45 seconds.
The centripetal force acting on the sphere is given by the equation:
F_c = m * (2πr / T)^2
Where:
F_c is the centripetal force,
m is the mass of the sphere,
r is the radius of the circular wire (15 cm = 0.15 m),
and T is the period of revolution (0.45 seconds).
The centripetal force is balanced by the gravitational force acting on the sphere:
F_g = m * g
Where:
F_g is the gravitational force,
m is the mass of the sphere,
and g is the acceleration due to gravity (approximately 9.8 m/s^2).
Setting these two equations equal to each other, we have:
m * (2πr / T)^2 = m * g
Simplifying the equation, we find the mass of the sphere cancels out:
(2πr / T)^2 = g
Next, we can solve for the angle up from the bottom at which the sphere will slide.
Since the only force acting on the sphere is the gravitational force, we can equate the component of the gravitational force along the wire (mg * cosθ) with the centrifugal force (m * (2πr / T)^2) to find the angle θ:
mg * cosθ = m * (2πr / T)^2
Canceling out the mass, we get:
g * cosθ = (2πr / T)^2
Simplifying further, we can solve for cosθ:
cosθ = (2πr / T)^2 / g
For this specific case, substituting the given values, we get:
cosθ = (2π * 0.15 / 0.45)^2 / 9.8
Using a calculator, we can compute the value of cosθ and find the angle θ:
cosθ ≈ 0.4391
θ ≈ cos^(-1)(0.4391)
Therefore, the angle up from the bottom at which the sphere will slide and maintain a steady height is approximately θ ≈ 64.9621 degrees.
(b) Now let's consider the second case where the period is 0.85 seconds.
Following the same equation and steps as in part (a), we can calculate the angle θ for this case as well.
cosθ = (2πr / T)^2 / g
Substituting the given values:
cosθ = (2π * 0.15 / 0.85)^2 / 9.8
Using a calculator, we can compute the value of cosθ and find the angle θ:
cosθ ≈ 0.2659
θ ≈ cos^(-1)(0.2659)
Therefore, the angle up from the bottom at which the sphere will slide and maintain a steady height is approximately θ ≈ 75.0521 degrees.
To solve this problem, we need to consider the forces acting on the sphere when it is spinning on the circular wire.
Let's begin with part (a):
(a) If the circular loop of wire is spun around its vertical diameter with a period of 0.45 seconds, at what angle up from the bottom will the sphere slide so that it stays at a steady height?
First, let's consider the forces acting on the sphere:
1. Gravitational force (mg): This force acts vertically downward and is equal to the mass of the sphere (m) multiplied by the acceleration due to gravity (g).
2. Centripetal force (Fc): When the sphere is spinning on the circular wire, it experiences a centripetal force directed towards the center of the circle. This force is equal to the mass of the sphere (m) multiplied by the radial acceleration (ar). The radial acceleration is given by the formula: ar = v^2 / r, where v is the linear velocity and r is the radius of the circular wire.
Since the sphere is sliding without friction on the wire, the centripetal force is provided by the normal force (N) acting on the sphere. The normal force has both a vertical and horizontal component.
Now, let's calculate the angle up from the bottom at which the sphere will slide so that it stays at a steady height:
1. Determine the normal force (N): The total normal force acting on the sphere can be split into a vertical component (Nv) and a horizontal component (Nh). In this case, the vertical component balances the gravitational force, so we have Nv = mg.
2. Calculate the horizontal component of the normal force (Nh): The horizontal component of the normal force is related to the centripetal force. Since the sphere slides without friction, Nh = Fc. Therefore, Nh = m * v^2 / r.
3. Find the angle up from the bottom: The angle up from the bottom can be determined by taking the inverse tangent of Nh / Nv. Therefore, the angle θ = arctan(Nh / Nv) = arctan(m * v^2 / (r * mg)).
Now, let's move on to part (b):
(b) Repeat the problem using a period of 0.85 seconds.
To solve this part, you will follow the same steps as above, but with a different period of 0.85 seconds. Calculate the angle θ using the formula θ = arctan(m * v^2 / (r * mg)), where v is determined as v = 2πr / T, with T being the new period.
Remember to convert any values to appropriate units for consistency, such as meters for distance and seconds for periods.