The rotational kinetic energy, KEr, is obtained from the translational kinetic energy, KEt = ½mv², by using the moment of inertia, I, for the mass and the rotational velocity, ω, for the translational velocity.
KEr = ½Iω²
The moment of inertia depends on the mass, the radius of rotation, and the shape of the object. For a mass at a radius r from the center of rotation, the moment of inertia is defined as I = mr². A solid rod of length d, rotated about an end has a moment of inertia of I = ⅓md². The moments of inertia for some other shapes, having radius r, rotated lengthwise through the center (like rolling down an incline) are:
| Solid cylinder: | I = ½mr² |
| Hollow cylinder: | I = mr² |
| Solid sphere: | I = ⅖mr² |
| Hollow sphere: | I = ⅔mr² |
The total kinetic energy of an object is obtained by adding the translational kinetic energy, KEt and the rotational kinetic energy, KEr. It is the total kinetic energy that must be used in the conservation of energy expression.
What is the final velocity of a hollow cylinder that starts from rest and rolls 3 m down a 30° smooth incline?
The conservation of energy expression would be:
KEit + KEir + PEi = KEft + KEfr + PEf + W
0 + 0 + mghi = ½mv² + ½Iω² + 0 + 0
mghi = ½mv² + ½(mr²)ω²
The mass cancels out, g = 9.8 m/sec², hi = (3 m)sin30 = 1.5 m, and ω = v/r.
(9.8 m/sec²)(1.5 m) = ½v² + ½(r²)(v/r)²
(9.8 m/sec²)(1.5 m) = ½v² + ½v² = v²
v = [(9.8 m/sec²)(1.5 m)]½ = 3.83 m/sec
Some time ago I developed the beginnings of a Flash Simulation that calculates this final speed and more for a hollow cylinder, a hollow sphere, a solid sphere, and for sliding. I have also just written an html5 version for a solid sphere rolling down a ramp, another one for a box sliding down a ramp and a third one that does several shapes like the old flash simulation.
| Linear | Rotational |
|---|---|
| v = vi + at | ω = ω² + αt |
| v² = vi² + 2as | ω² = ωi² + 2αθ |
| s = vit + ½at² | θ = ωit + ½αt² |
All angles are in radians and the rotational velocities are in radians/sec. When this is the case α = a/r, ω = v/r, and the linear distance s becomes the angular distance θ.
A wheel with a radius of 50 cm is rolling along a path at 8 m/sec when a mud puddle is encountered, causing the wheel to slow down and stop in 5 revolutions. How much time elapses while the wheel is stopping in the mud puddle?
ω = 0 (The wheel came to a stop.)
ωi = v/r = (8 m/sec)/(0.5 m) = 16 rad/sec
θ = (5 rev)(2π rad/rev) = 31.42 rad
α = (ω² -ωi²)/2θ = [0 - (16 rad/sec)²]/2(31.42 rad) = -4.07 rad/sec²
t = (ω - ωi)/α = (0 - 16 rad/sec)/(-4.07 rad/sec²) = 3.93 sec