Question

Consider a circular wheel with a mass m and a radius r. The moment of inertia about the center of the wheel is given by i = k m r², where k is a constant in the range between 0.5 ≤ k ≤ 1.0. A rope wraps around the wheel and a weight of mass 2m is attached to the end of this rope. At some moment, the weight is falling with a speed v. The total kinetic energy K of the system at this moment is given by?

1) K = 0.5 m v²
2) K = 2 m v²
3) K = 4 m v²
4) K = 8 m v²

Answer 1

The total kinetic energy K of the system consisting of a circular wheel and a falling weight will range between 1.5 * m * v^2 and 2 * m * v^2, depending on the specific value of k. The closest multiple-choice answer provided is K = 2 m v^2.

Step-by-step explanation:

To find the total kinetic energy K of the system which includes both the translational kinetic energy of the falling weight and the rotational kinetic energy of the wheel, we first calculate the translational kinetic energy of the weight using the formula Ktrans = 0.5 * mweight * v2, where mweight = 2m is the mass of the weight. Next, we determine the wheel's rotational kinetic energy using the formula Krot = 0.5 * I * ω2, where I is the moment of inertia of the wheel and ω is the angular velocity. Given that the moment of inertia I = k * m * r2 and the angular velocity ω can be expressed in terms of v as ω = v / r, we substitute these into the formula to get Krot = 0.5 * k * m * r2 * (v / r)2 = 0.5 * k * m * v2. Since both the falling weight and the wheel are parts of the same system, their kinetic energies add up to the total kinetic energy of the system: K = Ktrans + Krot = 0.5 * 2m * v2 + 0.5 * k * m * v2. Because k is between 0.5 and 1.0, the total kinetic energy K will range between 1.5 * m * v2 and 2 * m * v2. Therefore, the closest answer from the multiple choice given in the question would be 2) K = 2 m v2, though it's worth noting that the precise answer depends on the specific value of k.