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Chapter 8: Problem 2

What is the rotational inertia of a solid iron disk of mass \(49 \mathrm{kg},\) with a thickness of \(5.00 \mathrm{cm}\) and radius of \(20.0 \mathrm{cm}\) about an axis through its center and perpendicular to it?

Short Answer

Expert verified
Question: Calculate the rotational inertia of a solid iron disk with a mass of 49 kg, thickness of 5.00 cm, and radius of 20.0 cm. Answer: The rotational inertia of the solid iron disk is 0.980 kg × m^2.

Step by step solution

01

Write down the given values

We are given the following values: Mass of the disk (m) = 49 kg Thickness of the disk = 5.00 cm = 0.05 m (converted to meters) Radius of the disk (R) = 20.0 cm = 0.20 m (converted to meters)
02

Calculate the volume of the disk

To find the volume of the solid iron disk, we use the formula for the volume of a cylinder: Volume (V) = π × R^2 × thickness Plug in the values, and compute the volume. V = π × (0.20 m)^2 × 0.05 m ≈ 0.0062832 m^3
03

Use the formula for rotational inertia of a solid disk

The formula for rotational inertia (I) of a solid disk about an axis through its center and perpendicular to the disk is given by: I = 1/2 × m × R^2
04

Calculate the rotational inertia of the disk

Plug in the values of mass (m) and radius (R) into the formula for rotational inertia (I). I = 1/2 × 49 kg × (0.20 m)^2 ≈ 0.980 kg × m^2 The rotational inertia of the solid iron disk is 0.980 kg × m^2.

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Most popular questions from this chapter

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