Anyone who has spun a potter’s wheel is appreciative of the smooth motion of the flywheel upon which they form their clay, for without it the bowl they’re forming would display irregularities such as unattractive bumps. The flywheel’s smooth action comes as a result of Let’s begin by referring back to the formula for calculating
where Flywheels rotate about a fixed point rather than move in a straight line, but determining the amount of kinetic angular velocity and linear velocity. In fact, the amount of kinetic energy stored within it depends on how fast it rotates. For our example we’ll consider a spinning flywheel, which is basically a solid disc. For our illustrative purposes we’ll divide this disc into hypothetical parts, each having a mass
Part r from the flywheel’s center of rotation. As the flywheel spins, part _{A}A rides along with it at an angular velocity, ω, following a circular path, shown in green. It also moves at a linear velocity, v, shown in red. _{A}v represents the linear velocity of part _{A}A measured at any point tangent to its circular path. This linear velocity would become evident if part A were to become disengaged from the flywheel and cast into the air, whereupon its trajectory would follow a straight line tangent to its circular path. The linear and angular velocities of part
Next time we’ll use this equation to modify the basic kinetic energy formula so that it’s placed into terms that relate to a flywheel’s
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