The Physics
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Opus in profectus

# Rotational Energy

## Practice

### practice problem 1

A flywheel is a rotating mechanical device used to store mechanical energy. When attached to a combined electric motor-generator, flywheels are a practical way to store excess electric energy. Solar farms only generate electricity when it's sunny and wind turbines only generate electricity when it's windy. Combining energy sources like solar and wind with flywheel energy storage devices like a flywheel is one way to create a renewable energy system that is load balanced.

Given the energy storage flywheel described in the table, answer the following questions.
1. What is the mass of the flywheel?
2. What is the top angular speed of the flywheel?
3. For how long could a fully charged flywheel deliver maximum power before it needed recharging?
4. What is the average angular acceleration of the flywheel when it is being discharged?

#### solution

Pay attention to the units throughout this problem. It's a mix of SI units (kg/m3), SI units with prefixes (cm, kW), and acceptable non-SI units (h).

1. Use the definition of density…

 ρ = m V

and the volume of a cylinder…

V = πr2h

to find mass.

 m = ρ(πr2h)m = (7,850 kg/m3)π(0.50 m)2(0.60 m)m = 3,700 kg
2. Use the kinetic energy equation…

K = ½Iω2

and the moment of inertia of a cylinder…

I = ½MR2

to find angular speed.

 ω = 2 √ K R M
 ω = 2 √ (32,000 W)(3,600 s) 0.50 m 3,700 kg
 ω = 706 rad/s = 112 rpm
3. Use the definition of average power…

 P = ∆W ∆t

to find time

 ∆t = ∆W P
 ∆t = 32 kWh 8 W
 ∆t = 4 h

This is why the kilowatt-hour was invented. It makes some calculations more relatable.

4. Use the definition of angular acceleration to find angular acceleration.

 α = ∆ω ∆t
 α = −706 rad/s 4 × 3,600 s

### 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.

 translation rotation U0 = Kt + Kr mgh = ½mv2 + ½Iω2 mgs = ½m(Rω)2 + ½[½m(R2 + r2)]ω2 4mgs = 2mR2ω2 + mR2ω2 + mr2ω2

Clean it up a bit.

4gs = (3R2 + r22

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 = Rωv = (0.060 m)(103 rad/s)v = 6.17 m/s
3. angular acceleration (with a tiny modification)…

ω2 =  ω02 + 2α∆θ

ω2 =
 2α ⎛⎜⎝ s ⎞⎟⎠ R

α =
 Rω2 2s

2(3.0 m)

4. and translational acceleration…

 a = Rαa = (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. Straightforward. Do it.

 ∑F = mamg − T = maT = 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.

Magnify

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 =  Kt

Fs =  ½mv02

T =
 mv02 2s

T =  (0.50 kg)(10 m/s)2
2(2.5 m)
T =  10 N

### 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.

Magnify

2. Draw a velocity field diagram.

Magnify

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 = ½v2dmdK = ½ρv2dVdK = ½ρv2 dr rdθ h

Note that the infinitesimal volume isn't dx dy h (which looks like a box or a slab), it's 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 2π reye ⌠⎮⌡ dK = ½ρhv2max ⌠⎮⌡ ⌠⎮⌡ r ⎛⎜⎝ r ⎞2⎟⎠ dr dθ reye 0 0 0
 reye 2π K = ½ρhv2max ⎡⎡⎢⎢⎣⎣ θr4 ⎤⎥⎦ ⎤⎥⎦ 4reye2 0 0

K = ¼πρhv2maxr2eye

Then, outside the eye…

 V 2π R ⌠⎮⌡ dK = ½ρhv2max ⌠⎮⌡ ⌠⎮⌡ r ⎛⎜⎝ reye ⎞⎟⎠ dr dθ r 0 0 reye
 R 2π K = ½ρhv2max ⎡⎣ ⎡⎣ reyerθ ⎤⎦ ⎤⎦ 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 = 23R

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