University:
Santa Fe College
Course:
PHY 2004 | Applied Physics 1
Academic year:

2024
Views:

59

Pages:

6
Author:

Jabin F.

# Canonical Pendulum: 3 Here we have to find out tension in the string and angle made with vertical axis. where ω, g, m are given. Now radial EOM: -T sin θ = m(-ω²x) = -mω²l sin θ -T = mω²l T = mω²l As circle is horizontal, there is equilibrium in vertical direction. T cos θ = mg ⇒ mω²l cos θ = mg cos θ = g / (ω²l) ⇒ θ = cos⁻¹(g / (ω²l)) P-58 A hoop of radius a rotates with constant angular velocity ω about the vertical axis as shown in the figure. A bead of mass m can slide on the hoop without friction. If g < ω²a, at what angle θ apart from θ = 0 and θ = π is the bead stationary (i.e. dθ/dt = d²θ/dt² = 0)? Solution: Same as above, where a is in place of l and N is in place of T. θ = cos⁻¹(g / (ω²a)) Here all the forces are similar to forces in canonical pendulum. P-61 Consider a simple pendulum in 3-d space. It consists of a string length l = 20 cm and bob mass m = 15 kg, attached to it as shown in figure below. The g is 10 m/s². The pendulum is pulled in the x-z plane to a position where the string makes an angle θ = π/3 with the z-axis. It is then released with an angular velocity 2 radians per second so that the angle, the string makes with z-axis does not change with time? Solution: θ = π/3 ω = 2 rad/s g = 10 m/s² l = 20 cm = 0.2 m m = 15 kg To maintain constant height, the stuntman will only move relative in a circular path. Here we use cylindrical coordinates. x = r(θ) cos ϕ y = r(θ) sin ϕ z = a(x² + y²) Now radial equation of motion: ṁr - (m/r)θ̇² = mr̈ = -N sin θ vertical equilibrium: N cos θ = mg dividing by we get: tan θ = (ω²r/g)sin θ tan θ = ω²/g ω = √(g * tan θ) = √(10 * tan (π/3)) = √(10 * √3) = √(10√3) rad/s Q-40 A bead of mass M slides along a parabolic wire described by z = g(x² + y²). The wire rotates with angular velocity Ω about the z-axis. At what value of Ω does the bead maintain a constant non-zero height under the action of gravity along -z? Solution: Using cylindrical coordinates: x = r cos ϕ y = r sin ϕ z = g(x² + y²) = g(r²cos²ϕ + r²sin²ϕ) = gr² radial EOM: Fx = ma -N sin θ = M(-rω²) N sin θ = Mrω² vertical equilibrium: N cos θ = Mg dividing (1) by (2): tan θ = rω²/g Now tan θ can be obtained as the slope of tangent of surface z = gx²: tan θ = dz/dx = 4gx ω = √(g/4x) = (1/2)√(g/x)