Conservation of angular momentum

Problem:

A child of mass 25 kg stands at the edge of a rotating platform of mass 150 kg and radius 4.0 m.  The platform with the child on it rotates with an angular speed of 6.2 rad/s.  The child jumps off in a radial direction (i.e. has no tangential velocity in the inertial frame of an outside observer).  What happens to the angular speed of the platform?  Treat the platform as a uniform disk.  Give a quantitative explanation.

Solution:

• Concepts:
  Conservation of angular momentum
• Reasoning:
  The angular momentum of two interacting objects is constant.
• Details of the calculation:
  (I_platform + I_child)ω_i = I_platformω_f.
  I_platform = ½ 150kg (4m)² = 1200 kgm²,  I_child = 25kg (4m)² = 400 kgm².
  ω_f = 8.27 rad/s.

Problem:

A 60 kg woman stands at the rim of a horizontal turntable having a moment of inertia of 500 kgm², and a radius of 2 m.  The turntable is initially at rest and is free to rotate about a frictionless vertical axis through its center.  The woman then starts walking around the rim clockwise (as viewed from above the system) at a constant speed of 1.5 m/s relative to the Earth.
(a)  In what direction and with what angular speed does the turntable rotate?
(b)  How much work does the woman do to set herself and the turntable in motion?

Solution:

• Concepts:
  Conservation of angular momentum
• Reasoning:
  The system consists of the woman and the turntable.  No external torques act on the system, so the total angular momentum of the system is conserved.  It is zero before the woman starts to walk, and it is zero afterwards.
• Details of the calculation:
  (a)  When the women walks with angular velocity
  ω = -(v/r)k = -((1.5 m/s)/(2 m))k = -(0.75/s)k,
  her angular momentum is
  L = Iω = (-60 kg 4 m² 0.75/s)k = -(180 kgm²/s)k.
  The angular momentum of the turntable will be
  L = (180 kgm²/s)k,
  its angular velocity
  ω = (180 kgm²/s)k/(500 kgm²) = (0.36/s)k.  The turntable rotates counterclockwise.
  (b)  The kinetic energy of the women is ½Iω² = ½240 kgm²(0.75 s)² = 67.5 J.
  The kinetic energy of the turntable is ½Iω² = ½500 kgm²(0.36 s)² = 32.4 J.
  The work done by the women is W = (67.5 + 32.4) J = 99.9 J.

Problem:

A mass M is attached to the end of a string which passes through a hole in a frictionless horizontal plane.  Initially the mass moves on a circle of radius R with kinetic energy T₀.  The string is then slowly pulled so that the mass finally rotates on a circle of radius R/3.  How much work was done?

Solution:

• Concepts:
  Conservation of angular momentum
• Reasoning:
  We have a force directed towards the center of the orbit.  Therefore the torque is zero.
• Details of the calculation:
  Initial angular momentum:  L_i = mRv₀ = mR(2T₀/m)^½.
  Final angular momentum:  L_f = mRv_f /3 = m(R/3)(2T_f/m)^½.
  Conservation of angular momentum:  L_i = L_f,
  mR(2T₀/m)^½ = m(R/3)(2T_f/m)^½,  T₀ = T_f/9,  T_f = 9 T₀.  Work done: T_f - T₀ = 8 T₀.

Problem:

A small ball swings in a horizontal circle at the end of a cord of length L₁ which forms an angle θ₁ with the vertical.  Gravity is acting downward.  The cord is slowly shortened by pulling it through a hole in its support until the free length is L₂ and the ball is moving at an angle θ₂ from the vertical. 
(a)  Derive a relation between L₁, L₂, θ₁, and θ₂. 
(b)  If L₁ = 50 cm, θ₁ = 5^o, and L₂ = 30 cm, find θ₂.

Solution:

• Concepts:
  Conservation of angular momentum
• Reasoning:
  The applied force produces no torque.
• Details of the calculation:
  
  
  Initial motion:  T₁cosθ₁   = mg,  T₁sinθ₁  = mv₁²/r = mv₁²/(L₁sinθ₁), 
  v₁² = gL₁sin²θ₁/cosθ₁.
  Final motion :  v₂² = gL₂sin²θ₂/cosθ₂.
  Conservation of angular momentum: mL₁sinθ₁v₁ = mL₂sinθ₂v₂,  L₁²sin²θ₁v₁² = L₂²sin²θ₂v₂².
  Therefore L₁³sin³θ₁tanθ₁ = L₂³sin⁴θ₂tanθ₂.
  (b)  Small angle approximation
  L₁³θ₁⁴ = L₂³θ₂⁴,  θ₂⁴ = 2.68*10^-4,  θ₂ = 0.128 rad = 7.33^o.

Problem:

A point mass m attached to the end of a string revolves in a circle of radius R on a frictionless table at constant speed with initial kinetic energy E₀.  The string passes through a hole in the center of the table and the string is pulled down until the radius of the circle is ½ of its initial value. Assuming no external torque acts on the system, how much work is done?

Solution:

• Concepts:
  Conservation of angular momentum
• Reasoning:
  A radial force does not exert a torque about the origin.
• Details of the calculation:
  E₀ = ½mR²ω², L = mR²ω = (2mE₀)^½R.
  After the radius has been reduced to ½R, L = (2mE_f)^½R/2. 
  (2mE₀)^½R = (2mE_f)^½R/2.  4E₀ = E_f.  The work done on the system is 3E₀.
  (Or, E₀ = ½mv², L = mvR = mv_fR/2.  So v_f = 2v, E_f = ½mv_f² = 4E₀.  W = 4E₀ - E₀ = 3E₀.)

Problem:

A solid sphere toy globe of mass M and radius R rotates freely without friction with an initial angular velocity ω₀.  A bug of mass m starts at one pole N and travels with constant speed v to the other pole S along a meridian in time T.  The axis of rotation of the globe is held fixed.  Show that during the time the bug is traveling the globe rotates through an angle
Δθ = (πω₀R/v) (2M/(2M + 5m))^1/2.

Useful integral: ∫₀^2πdx/(a + b cosx) = 2π/(a² - b²)^1/2,  (a² > b²).

Solution:

• Concepts:
  Conservation of a component of the angular momentum
• Reasoning:
  The axis of rotation is held fixed, so external forces cannot be excluded.  But holding an axis fixed cannot exert a torque about this axis.  So the component of angular momentum about the axis of rotation is conserved.
• Details of the calculation:
  
  L_z =  Iω = I₀ω₀ = constant.
  I₀ = (2/5)MR²,  I = I₀ + mR²sin²φ.
  I(t)ω(t) = [(2/5)MR² + mR²sin²φ](dθ/dt) = (2/5)MR²ω₀.
  constant speed:  φ = vt/R.
  (dθ/dt) = (2/5)MR²ω₀/[(2/5)MR² + mR²sin²(vt/R)].
  Δθ =  ∫₀^Tdt (2/5)MR²ω₀/[(2/5)MR² + mR²sin²(vt/R)].
  Change variables:  x = 2vt/R,  sin²(x/2) = ½(1 - cosx), since ∫dx/(a + b cosx) is an integral found in integral tables and is given here.. 
  Use vT/R = π.
  Δθ =  (Rω₀/v)∫₀^2πdx/[2 + (5m/M) sin²(x/2)]
  = (Rω₀/v)∫₀^2πdx/[2 + (5m/(2M))(1 - cosx)]
  = (πRω₀/v)(2M/(2M + 5m))^1/2.
  As m --> 0, Δθ --> πω₀R/v = ω₀T, which corresponds to the free rotation of the globe with angular speed ω₀.

Problem:

A uniform spherical planet of radius a revolves around the sun in a circular orbit of radius r₀ and angular velocity ω₀.  It rotates about its axis with angular velocity Ω₀ (period T₀) normal to the plane of the orbit.  Due to tides raised on the planet by the sun, its angular velocity of rotation is decreasing.  Find an expression which gives the orbital radius r as a function of the angular velocity Ω of rotation and the parameters r₀ and T₀ at any later or earlier time.