The Physics
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Opus in profectus

Rotational Energy

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Practice

practice problem 1

Write something.

solution

Answer it.

practice problem 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

solution

  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 + r22
    2 2
    4mgs =  2mR2ω2  +  mR2ω2 + mr2ω2  
     
    Solve for angular speed and input numbers.
     
    ω = 
    4gs ½
       
    3R2 + r2  
    ω = 
    4(9.8 m/s2)(3.0 m) ½
    3(0.060 m)2 + (0.018 m)2
    ω = 103 rad/s = 16.4 rev/s  
     
  2. Use basic formulas to compute the translational speed…
     
    v = Rω
    v = (0.060 m)(103 rad/s)
    v = 6.17 m/s
  3. …angular acceleration (with a tiny modification)…
     
    ω2 =  ω02 + 2αΔθ  
     
    ω2 =  2α 
    s
       
    R  
    α =  Rω2      
    2s    
    α =  (0.060 m)(103 rad/s)2  
    2(3.0 m)
    α = 106 rad/s2  
     
  4. …and translational acceleration…
     
    a = 
    a = (0.060 m)(106 rad/s2)
    a = 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)
    T = (0.200 kg)(9.8 m/s2 − 6.34 m/s2)
    T = 0.691 N 

practice problem 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.

Using energy considerations determine…

  1. the tension T in the string
  2. something
  3. something else
  4. maybe something else

solution

  1. Pulling on the string does work on the top, destroying its initial translational kinetic energy.
     
    W =  ΔKt  
     
    Fs =  1  mv02    
    2
    T =  mv02    
    2s
    T =  (0.50 kg)(10 m/s)2  
    2(2.5 m)
    T =  10 N  
     
  2. answer
  3. answer
  4. answer

practice problem 4

The simplest mathematical models of hurricanes and typhoons (collectively known as tropical cyclones) describe a cylindrical mass of rotating air with no updrafts, downdrafts, or turbulence. In these vortex models, the air in a central region called the eye is often assumed to rotate as if it was one solid piece of material — slowest at the center and fastest at the outer edge or eye wall. (The eye wall, not the center, is the region of maximal wind speed in a hurricane.) Beyond the eye wall, wind speeds decay away according to a simple power law.

Here's an example of a vortex model of a hurricane with an outer region described by an inverse square root power law.

v(r) = 



 vmax 
r
0 ≤  r  ≤ reye
reye
 vmax 
reye ½
  r  ≥ reye
r

Where…

v(r) =  tangential wind speed
vmax =  wind speed at eye wall
r =  distance from center of hurricane
reye =  radius of eye wall

and…

h =  height of hurricane
R =  radius of hurricane (R > reye)
ρ =  average density of air

Given this model…

  1. Graph tangential wind speed as a function of radius.
  2. Draw a velocity field diagram.
  3. Derive an expression for the total kinetic energy of a storm.
  4. Determine the total kinetic energy of a tropical cyclone 500 km in diameter, 10 km tall, with an eye 10 km in diameter and peak winds speeds of 140 km/h. (Assume the average density of the air is 0.9 kg/m3.)

solution

  1. Graph tangential wind speed as a function of radius.

  2. Draw a velocity field diagram.

  3. Derive an expression for the total kinetic energy of a storm. Start with the definition of kinetic energy.

    K = ½mv2

    Break the storm up into little pieces and integrate the contributions to the total energy budget that each piece makes.

    dK = ½v2dm = ½ρv2dV = ½ρv2 dr rdθ h

    Note that the infinitesimal volume isn't dx dy h (which looks like a box or a slab), it's a dr rdθ h (which looks like an arch or a fingernail). Since this vortex model has two parts to it (inside and outside the eye) and the integral has two infinitesimals (one radial, one angular), we'll be doing four integrals. The integrals are all easy, but there are a lot of them. First, inside the eye…

    V reye

    dK = ½ρhv2max

    r 
    r 2
     dr dθ
    reye
    0 0 0
    reye
    K = ½ρhv2max 
     
    θr4

    4r2eye
    0 0
    K = ¼πρhv2maxr2eye  
     

    Then, outside the eye…

    V   R  

    dK = ½ρhv2max

    r 
    reye
     dr dθ
    r
    0   0 reye  
    R
      K = ½ρhv2max 
     
    reye

     
    reye 0
      K = πρhv2maxreye(R − reye)
     

    Combine the two…

     K = πρhv2maxreye(R − ¾reye

    This equation says that the total kinetic energy of a tropical cyclone…

    • is proportional to the square of the maximum wind speed, which agrees nicely with the basic equation of kinetic energy.
    • is directly proportional to its radius, which I find somewhat counter intuitive. I know that energy increases with size, but I silently suspected that size would be determined by area. Our analysis shows, however, that in this model, size is determined by radius.
    • increases as the radius of the eye increases, which I seem to remember hearing is true and now I see is true for this vortex model. Well, true up to a point. The equation we just derived is a quadratic function of reye and has a maximum value when…
       
      d  πρhv2maxreye(R − ¾reye) = 0
      dreye
        R − 32reye = 0
       
        reye = ⅔R
       
       
      A tropical cyclone that was two-thirds eye is unheard of (two-thirds measured along the radius or diameter). Your typical cyclone has an overall diameter measured in hundreds of kilometers and an eye diameter measured in tens of kilometers.
  4. Plug and chug. Watch out for an obvious mistake. Don't confuse diameter with radius.
       
    K =  πρhv2maxreye(R − ¾reye)
    K =  π(0.9 kg/m3)(10,000 m)(140,000 m/3,600 s)2(5,000 m)(250,000 m − ¾ 5,000 m)
    K =  5 × 1016 J