A particle of mass m1 is tide to a string and describes a circumference on a horizontal plane. The string (of negligible mass)passes through a hole that is in the center of the plane and goes down vertically.
At the lower end is tied another particle of m2. If the first particle's movement describes a uniform circle, with angular velocity w, determine the radius R of the circle to which the system is in balance
To determine the radius R of the circle to which the system is in balance, we need to consider the forces acting on the system.
Let's break down the forces acting on the particles individually:
1. Particle 1:
- Tension in the string: T1
- Centripetal force: Fc1 = m1 * w^2 * R, directed towards the center of the circle
2. Particle 2:
- Tension in the string: T2 = T1
- Weight: m2 * g, directed downward
At equilibrium, the system will have no net force acting on it vertically or horizontally.
Vertically:
- Tension forces in the string cancel out the weight of particle 2:
T2 = m2 * g
Horizontally:
- There are no net forces acting on particle 1 since it moves in a uniform circle.
Therefore, the centripetal force Fc1 is balanced by the horizontal component of T2:
Fc1 = T2 * cos(θ)
m1 * w^2 * R = m2 * g * cos(θ)
We need to find the relationship between θ and R.
To do that, we'll need to consider the geometry of the system.
The string forms a right-angled triangle with the vertical downward direction and the radius R as its hypotenuse.
In the triangle:
- The adjacent side is R (the horizontal component of T2).
- The hypotenuse is the length of the string, which is the sum of the vertical component of T2 (m2 * g * sin(θ)) and the length of the string below particle 2 (let's call it h).
Using Pythagorean's theorem, we have:
R^2 = h^2 + (m2 * g * sin(θ))^2
Now, let's simplify and solve for R:
1. Substitute h = R * cos(θ):
R^2 = (R * cos(θ))^2 + (m2 * g * sin(θ))^2
R^2 = R^2 * cos^2(θ) + m2^2 * g^2 * sin^2(θ)
2. Divide both sides by R^2 (assuming R is not zero):
1 = cos^2(θ) + (m2^2 * g^2 * sin^2(θ)) / R^2
3. Rearrange the equation to isolate R:
(m2^2 * g^2 * sin^2(θ)) / R^2 = 1 - cos^2(θ)
(m2^2 * g^2 * sin^2(θ)) / R^2 = sin^2(θ)
(m2^2 * g^2) / R^2 = 1
R^2 = (m2^2 * g^2)
Finally, taking the square root of both sides, we get:
R = m2 * g
Therefore, the radius R of the circle to which the system is in balance is equal to m2 multiplied by the acceleration due to gravity (g).