# Rotational Energy

## Practice

### practice problem 1

characteristic | value |
---|---|

shape | solid cylinder |

material | 4340 steel |

density | 7850 kg/m^{3} |

diameter | 100 cm |

height | 60 cm |

max. energy | 32 kWh |

max. power | 8 kW |

dc voltage | 800 V |

Given the energy storage flywheel described in the table, answer the following questions.

- What is the mass of the flywheel?
- What is the top angular speed of the flywheel?
- For how long could a fully charged flywheel deliver maximum power before it needed recharging?
- 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/m^{3}), SI units with prefixes (cm, kW), and acceptable non-SI units (h).

Use the definition of density…

ρ = *m**V*and the volume of a cylinder…

*V*= π*r*^{2}*h*to find mass.

*m*= ρ(π*r*^{2}*h*)*m*= (7,850 kg/m^{3})π(0.50 m)^{2}(0.60 m)*m*= 3,700 kgUse the kinetic energy equation…

*K*= ½*I*ω^{2}and the moment of inertia of a cylinder…

*I*= ½*MR*^{2}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 Use the definition of average power…

*P*=∆ *W*∆ *t*to find time

∆ *t*=∆ *W**P*∆ *t*=32 kWh 8 W ∆ *t*= 4 hThis is why the kilowatt-hour was invented. It makes some calculations more relatable.

Use the definition of angular acceleration to find angular acceleration.

α = ∆ω ∆ *t*α = −706 rad/s 4 × 3,600 s α = −0.049 rad/s ^{2}

### practice problem 2

*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…

- the final angular speed of the roll
- the final translational speed of roll
- the angular acceleration of the roll
- the translational acceleration of the roll
- the tension in the sheets

#### solution

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 (*R*ω). Replace the moment of inertia (*I*) with the equation for a hollow cylinder.translation rotation *U*_{0}=*K*_{t}+ *K*_{r}*mgh*=½ *mv*^{2}+ ½ *I*ω^{2}*mgs*=½ *m*(*R*ω)^{2}+ ½[½ *m*(*R*^{2}+*r*^{2})]ω^{2}4 *mgs*=2 *mR*^{2}ω^{2}+ *mR*^{2}ω^{2}+*mr*^{2}ω^{2}Clean it up a bit.

4

*gs*= (3*R*^{2}+*r*^{2})ω^{2}Solve for angular speed and input numbers.

ω = ⎛

⎜

⎝4 *gs*⎞ ^{½}

⎟

⎠3 *R*^{2}+*r*^{2}ω = ⎛

⎜

⎝4(9.8 m/s ^{2})(3.0 m)⎞ ^{½}

⎟

⎠3(0.060 m) ^{2}+ (0.018 m)^{2}ω = 103 rad/s = 16.4 rev/s Use basic formulas to compute the translational speed…

*v*=*R*ω*v*= (0.060 m)(103 rad/s)*v*= 6.17 m/sangular acceleration (with a tiny modification)…

ω ^{2}=ω _{0}^{2}+ 2α∆θω ^{2}=2α ⎛

⎜

⎝*s*⎞

⎟

⎠*R*α = *R*ω^{2}2 *s*α = (0.060 m)(103 rad/s) ^{2}2(3.0 m) α = 106 rad/s ^{2}and translational acceleration…

*a*=*R*α*a*= (0.060 m)(106 rad/s^{2})*a*= 6.34 m/s^{2}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*=*ma**mg*−*T*=*ma**T*=*m*(*g*−*a*)*T*= (0.200 kg)(9.8 m/s^{2}− 6.34 m/s^{2})*T*= 0.691 N

### practice problem 3

*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

*v*

_{0}= 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…

- the tension
*T*in the string - something
- something else
- maybe something else

#### solution

- Pulling on the string does work on the top, destroying its initial translational kinetic energy.
*W*=∆ *K*_{t}*Fs*=½ *mv*_{0}^{2}*T*=*mv*_{0}^{2}2 *s**T*=(0.50 kg)(10 m/s) ^{2}2(2.5 m) *T*=10 N - answer
- answer
- answer

### practice problem 4

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) = |
⎧ ⎪ ⎨ ⎪ ⎩ |
v _{max} |
⎛ ⎜ ⎝ |
r |
⎞ ⎟ ⎠ |
0 ≤ | r |
≤ r_{eye} |

r_{eye} |
||||||||

v _{max} |
⎛ ⎜ ⎝ |
r_{eye} |
⎞^{½}⎟ ⎠ |
r |
≥ r_{eye} |
|||

r |

where…

v(r) = |
tangential wind speed |

v = _{max} |
wind speed at eye wall |

r = |
distance from center of hurricane |

r = _{eye} |
radius of eye wall |

and…

h = |
height of hurricane |

R = |
radius of hurricane (R > r)_{eye} |

ρ = | average density of air |

Given this model…

- Graph tangential wind speed as a function of radius.
- Draw a velocity field diagram.
- Derive an expression for the total kinetic energy of a storm.
- 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/m
^{3}.)

#### solution

Graph tangential wind speed as a function of radius.

Draw a velocity field diagram.

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

*K*= ½*mv*^{2}Break the storm up into little pieces and integrate the contributions to the total energy budget that each piece makes.

*dK*= ½*v*^{2}*dm**dK*= ½ρ*v*^{2}*dV**dK*= ½ρ*v*^{2}*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 *r*_{eye}⌠

⎮

⌡*dK*= ½ρ*hv*^{2}_{max}⌠

⎮

⌡⌠

⎮

⌡*r*⎛

⎜

⎝*r*⎞ ^{2}

⎟

⎠*dr dθ**r*_{eye}0 0 0 *r*_{eye}2π *K*= ½ρ*hv*^{2}_{max}⎡⎡

⎢⎢

⎣⎣θ *r*^{4}⎤

⎥

⎦⎤

⎥

⎦4 *r*_{eye}^{2}0 0 *K*= ¼πρ*hv*^{2}_{max}r^{2}_{eye}Then, outside the eye…

*V*2π *R*⌠

⎮

⌡*dK*= ½ρ*hv*^{2}_{max}⌠

⎮

⌡⌠

⎮

⌡*r*⎛

⎜

⎝*r*_{eye}⎞

⎟

⎠*dr dθ**r*0 0 *r*_{eye}*R*2π *K*= ½ρ*hv*^{2}_{max}⎡

⎣⎡

⎣*r*_{eye}rθ⎤

⎦⎤

⎦*r*_{eye}0 *K*= πρ*hv*^{2}(_{max}r_{eye}*R − r*)_{eye}Combine the two…

*K*= πρ*hv*(^{2}_{max}r_{eye}*R*− ¾*r*)_{eye}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

*r*and has a maximum value when…_{eye}*d*πρ *hv*^{2}(_{max}r_{eye}*R*− ¾*r*) = 0_{eye}*dr*_{eye}*R*−^{3}_{2}*r*= 0_{eye}*r*=_{eye}^{2}_{3}*R*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.

Plug and chug. Watch out for an obvious mistake. Don't confuse diameter with radius.

*K*= πρ*hv*^{2}(_{max}r_{eye}*R*− ¾*r*)_{eye}*K*= π(0.9 kg/m^{3})(10,000 m) (140,000 m/3,600 s) ^{2}(5,000 m) (250,000 m − ¾ 5,000 m) *K*= 5 × 10^{16}J