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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 脳 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) = 蟺 脳 R^2 脳 thickness Plug in the values, and compute the volume. V = 蟺 脳 (0.20 m)^2 脳 0.05 m 鈮� 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 脳 m 脳 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 脳 49 kg 脳 (0.20 m)^2 鈮� 0.980 kg 脳 m^2 The rotational inertia of the solid iron disk is 0.980 kg 脳 m^2.

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Most popular questions from this chapter

A 46.4 -N force is applied to the outer edge of a door of width $1.26 \mathrm{m}$ in such a way that it acts (a) perpendicular to the door, (b) at an angle of \(43.0^{\circ}\) with respect to the door surface, (c) so that the line of action of the force passes through the axis of the door hinges. Find the torque for these three cases.
The string in a yo-yo is wound around an axle of radius \(0.500 \mathrm{cm} .\) The yo-yo has both rotational and translational motion, like a rolling object, and has mass \(0.200 \mathrm{kg}\) and outer radius \(2.00 \mathrm{cm} .\) Starting from rest, it rotates and falls a distance of \(1.00 \mathrm{m}\) (the length of the string). Assume for simplicity that the yo-yo is a uniform circular disk and that the string is thin compared to the radius of the axle. (a) What is the speed of the yo-yo when it reaches the distance of $1.00 \mathrm{m} ?$ (b) How long does it take to fall? [Hint: The transitional and rotational kinetic energies are related, but the yo-yo is not rolling on its outer radius.]
A bicycle wheel, of radius \(0.30 \mathrm{m}\) and mass \(2 \mathrm{kg}\) (concentrated on the rim), is rotating at 4.00 rev/s. After 50 s the wheel comes to a stop because of friction. What is the magnitude of the average torque due to frictional forces?
Verify that \(\frac{1}{2} I \omega^{2}\) has dimensions of energy.

A man is doing push-ups. He has a mass of \(68 \mathrm{kg}\) and his center of gravity is located at a horizontal distance of \(0.70 \mathrm{m}\) from his palms and \(1.00 \mathrm{m}\) from his feet. Find the forces exerted by the floor on his palms and feet.
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