rotational kinetic energy explained
A spinning object stores energy too. A top or a flywheel that merely spins in place still takes work to stop, and that energy is the rotational kinetic energy ½Iω². What changes it is the work done by a torque, W = Mθ, torque times the angle turned. A rolling object carries both the translational kinetic energy of its moving center and the rotational kinetic energy of its spin, which is why objects rolling down from the same height arrive at different speeds depending on their shape. This lesson revisits rotation through energy and ties the whole of dynamics together.
Rotation has kinetic energy too, KE = ½ I ω². It has the same form as ½mv² in straight-line motion, with the moment of inertia I in place of the mass m and the angular velocity ω in place of the velocity v. Increase ω with the slider and the disk spins faster while the gold bar, its rotational KE, swells with the square of ω. Double the angular velocity and the rotational kinetic energy becomes four times larger. That is why a heavy flywheel with a large moment of inertia, kept spinning fast, can store a lot of energy and release it when needed.
Just as work in straight-line motion was W = F d, in rotation the work done by a torque turning something through an angle θ is W = M θ. Increase the angle turned with the slider and the gold arc widens, while under a constant torque the work in the green bar builds up in proportion to the angle. Torque plays the role of force and angle the role of distance in rotation, so the formula keeps its shape. Divide this work by the time taken and you get power, which in rotation is P = M ω, torque times angular velocity. That relationship is why a car engine's output is described by its torque and its revolutions together.
A rolling object carries two kinds of kinetic energy at once: translational ½mv², from its center moving forward, and rotational ½Iω², from its spin. When it rolls without slipping, ω = vr, so the rotational share becomes ½βmv², and the total adds up to KE = ½mv²(1+β). The disk in the figure has β of ½; change the speed with the slider and the blue translational segment and the orange rotational segment grow together, always in the same ratio. So an object rolling at a given speed has more kinetic energy than one merely sliding at that speed.
Rolling down a slope, the potential energy at the top, mgh, turns entirely into rolling kinetic energy ½mv²(1+β) at the bottom. The upper green bar is the potential energy at the start and the lower bar the kinetic energy at the finish, and the two bars always have the same length. Change the height h with the slider and both bars grow and shrink together, the orange rotational share always taking up the same proportion of the lower bar. Because part of the energy goes into rotation, the arrival speed from the same height is lower than for an object sliding down without friction.
So objects starting from the same height arrive at different speeds depending on their shape. The arrival speed is v = √(2gh(1+β)), so the sphere, whose small β means rotation takes less of its energy, is fastest and the hoop slowest. Switch shapes with the buttons and the green bar shows that shape's arrival speed, with the tick for the chosen shape highlighted among the three ticks on the bar. The sphere's tick sits farthest right and the hoop's farthest left, so the order is clear at a glance. Mass and radius do not affect the speed; only the shape, how the mass is spread out, sets the order.