Final answer:
To find the final rotational kinetic energy of an electric sander's disk after the rotation rate decreases by 20%, we calculate it using the formula KErot = (1/2)Iω², where I am (1/2)mr² and ω is the final angular velocity in rad/s. After solving the given values, we find that the correct answer is KErot = 0.105 J.
Step-by-step explanation:
The subject of the question is Physics, and it is a High School level problem concerning rotational motion and kinetic energy. To solve for the final rotational kinetic energy of a rotating disk when its rotation rate decreases by 20%, we need to use the formula for rotational kinetic energy, which is KErot = (1/2)Iω², where I is the moment of inertia and ω (omega) is the angular velocity. The moment of inertia of a disk is I = (1/2)mR², with m being the mass and R being the radius. We are given that the original rotation rate is 15 rev/s which decreases by 20%, thus the final rotation rate is 80% of the initial rate, or 12 rev/s. Converting this to radians per second, we obtain ω = 12 rev/s × (2π rad/rev) = 24π rad/s. Substituting the values into the kinetic energy formula, we find:
KErot = (1/2) × (1/2) × 0.7 kg × (0.1 m)² × (24π rad/s)²
Solving this, we get:
KErot = 0.105 J
The correct option is (b) 0.105 J. This is the final rotational kinetic energy of the disk after a decrease in rotation rate due to the rough wooden wall.