Rotational Energy

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Discussion

Rotational kinetic energy

Translational and Rotational Quantities Compared
	translational		connections		rotational
work-energy	W =	∫ F · ds			W =	∫ τ · dθ
kinetic energy	K =	½ mv2			K =	½ Iω2
potential energy	U = F(x) =	− ∫ F · ds − dU / dx			U = τ(θ) =	− ∫ τ · dθ − dU / dθ
power	P =	F · v			P =	τ · ω

For a system of point bodies

K =	1	∑ mivi2 =	1	∑ ri2mi	vi2	=	1	Iω2
	2		2		ri2		2

For an extended body

K =	1	⌠ ⌡	v2 dm =	1	⌠ ⌡	v2	r2 dm =	1	ω2I =	1	Iω2
	2			2		r2		2		2

Moment of inertia

I = ∑r2m =

⌠ ⌡

r2 dm = ρ

⌠ ⌡

r2 dV

Angular work

W =

⌠ ⌡

F · ds =

⌠ ⌡

F · (dθ × r) =

⌠ ⌡

dθ · (r × F) =

⌠ ⌡

dθ · τ =

⌠ ⌡

τ · dθ

and so on

Summary

• bullet

Problems

practice

1. Write something.
   • Answer it.
2. A roll of toilet paper is held by the first piece and allowed to unfurl as shown in the diagram to the right. The roll has an outer radius R = 6.0 cm, an inner radius r = 1.8 cm, a mass m = 200 g, and falls a distance s = 3.0 m. Assuming the outer diameter of the roll does not change significantly during the fall, determine …
   1. the final angular speed of the roll
   2. the final translational speed of roll
   3. the angular acceleration of the roll
   4. the translational acceleration of the roll
   5. the tension in the sheets
   There are two general approaches to this problem: using conservation of energy (see below) and Newton's second law (see the section on rotational dynamics).
   1. The potential energy of the roll at the top becomes kinetic energy in two forms at the bottom. Replace the translational speed (v) with its rotational equivalent (Rω), replace the moment of inertia (I) with the equation for a hollow cylinder (see below), and clean it up a bit.

      	translation		rotation
      U0 =	Kt	+	Kr
      mgh =	1   mv2 2	+	1   Iω2 2
      mgs =	1   m(Rω)2 2	+	1     1   m(R2 + r2)ω2 2 2
      4mgs =	2mR2ω2	+	mR2ω2 + mr2ω2

      Solve for angular speed and input numbers.

      ω =	⎛ ⎜ ⎝	4gs	⎞½ ⎟ ⎠	=	⎛ ⎜ ⎝	4(9.8 m/s2)(3.0 m)	⎞½ ⎟ ⎠	= 103 rad/s = 16.4 rev/s
      		3R2 + r2				3(0.060 m)2 + (0.018 m)2

   2. Use basic formulas to compute the translational speed …

      v = Rω = (0.060 m)(103 rad/s) = 6.17 m/s

   3. … angular acceleration (with a tiny modification) …

      ω2 =	ω02 + 2αΔθ = 2α	⎛ ⎝	s R	⎞ ⎠
      α =	Rω2  =  (0.060 m)(103 rad/s)2  = 106 rad/s2 2s 2(3.0 m)

   4. … and translational acceleration …

      a = Rα = (0.060 m)(106 rad/s2) = 6.34 m/s2

   5. To compute the tension begin with Newton's second law of motion (let down be positive), work a little bit of algebra, substitute numbers, and compute. Very straightforward.

      ∑F =	ma
      mg − T =	ma
      T =	m(g − a) = (0.200 kg)(9.8 m/s2 − 6.34 m/s2) = 0.691 N

3. The top shown below consists of a cylindrical spindle of negligible mass attached to a conical base of mass m = 0.50 kg. The radius of the spindle is r = 1.2 cm and the radius of the cone is R = 10 cm. A string is wound around the spindle. The top is thrown forward with an initial speed of v₀ = 10 m/s while at the same time the string is yanked backward. The top moves forward a distance s = 2.5 m, then stops and spins in place.

   [magnify]

   Using energy considerations determine …
   1. the tension T in the string
   2. something
   3. something else
   4. maybe something else
   Solutions …
   1. Pulling on the string does work on the top, destroying its initial translational kinetic energy.

      W =	ΔKt
      Fs =	1	mv02
      	2
      T =	mv02		=	(0.50 kg)(10 m/s)2	= 10 N
      	2s			2(2.5 m)

   2. answer
   3. answer
   4. answer
4. Write something different.
   • Answer it.

numerical

1. problems

Resources

• flywheels
  • Bolund B, Bernhoff H, Leijon M. Flywheel energy and power storage systems. Renewable and Sustainable Energy Reviews. Vol. 11 No. 2 (February 2007): 238-258
  • Designing Safer Flywheels, Steven Ashley, Mechanical Engineering, Vol. 118, No. 11 (November 1996)
  • Distributed Energy
    • A Primer of Flywheel Technology, Henry Vere, (December 2007)
    • The Spin on Flywheels, Henry Vere, (December 2007)
  • Energy Storage Solutions, Beacon Power
  • Flywheels, Research Institute for Sustainable Energy (RISE), Murdoch University
  • From Child's Toy to ISS: Flywheels Hold the Power, Office of Biological and Physical Research, NASA

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