Question

A 50.0-kg skater is approximated as a solid cylinder that has a 50 cm diameter. Calculate the moment of inertia (in kg-m²) of the skater about the axis of the cylinder (which is the same as the axis of bilateral symmetry). Express your answer correct to 3 decimal places.

Answer 1

The moment of inertia of a 50.0-kg skater approximated as a solid cylinder about the cylinder's axis is 1.5625 kg·m².

Step-by-step explanation:

The student asked to calculate the moment of inertia of a 50.0-kg skater approximated as a solid cylinder with a diameter of 50 cm. To find the moment of inertia (I) for a solid cylinder about its axis, we use the formula I = (1/2)MR², where M is the mass of the cylinder, and R is the radius. Since the skater is 50.0 kg and the radius is half of the diameter (50 cm / 2 = 25 cm = 0.25 m), we can plug in the values into the formula to get I = (1/2)(50.0 kg)(0.25 m)² = (1/2)(50.0)(0.0625 kg·m²) = 1.5625 kg·m².