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------------------------------------------------ A solid cylinder of mass 20 kg rotates about its axis with an angular speed of 100 rad/s. The radius of the cylinder is 0.25 m. What is the kinetic energy associated with the rotation of the cylinder?

A. 3125 J
B. 725 J
C. 31.25 J
D. 7.25 J

The kinetic energy associated with the rotation of the cylinder is 3125 J, found by using the formula for rotational kinetic energy and substituting the values for mass, radius, and angular speed.

Explanation:

The student has asked about the kinetic energy of a rotating solid cylinder with given mass, angular speed, and radius. The formula for rotational kinetic energy (KErot) is KErot = 1/2 I ω2, where I is the moment of inertia of the cylinder and ω is its angular speed. For a solid cylinder, I = 1/2 MR2, where M is the mass and R is the radius of the cylinder. Substituting the given values

M = 20 kg, R = 0.25 m, and ω = 100 rad/s into the equation,

we calculate the kinetic energy.

First, find the moment of inertia:

I = 1/2 (20 kg) (0.25 m)2 = 0.625 kg·m2

Then, calculate the kinetic energy:

KErot = 1/2 (0.625 kg·m2) (100 rad/s)2 = 1/2 (0.625) (10000) = 3125 J