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Chapter 9: Problem 18

Calculate the rotational inertia of a meter stick, with mass \(0.56 \mathrm{~kg}\), about an axis perpendicular to the stick and located at the 20 -cm mark.

Short Answer

Expert verified
The moment of inertia (or rotational inertia) of the meter stick about an axis at the 20 cm mark is \(0.2371 kg*m^2\)

Step by step solution

01

Establish the mathematical representation

The mass of the meter stick is 0.56 kg and its length is 1 meter or 100 cm. The axis of rotation is 20 cm from one end of the meter stick, so it is 50 - 20 = 30 cm away from the center of mass. Using the given mass and length, calculate the inertia around the center of mass using the formula for a rod rotating around its end, \(I_{cm} = \frac{1}{3}mL^2\).
02

Calculate the inertia around the center of mass

Plug the values into the formula to get \(I_{cm} = \frac{1}{3}*0.56 kg*(1 m)^2 = 0.1867 kg*m^2\).
03

Use the Parallel Axis Theorem to calculate the inertia about the 20 cm mark

Now that the inertia around the center of mass is known, use the Parallel Axis Theorem, \(I = I_{cm} + md^2\), to calculate the inertia about the new axis located 30 cm from the center of mass. Convert 30 cm to meters by dividing by 100. So, \(d = 30 cm / 100 = 0.3 m\). Now substitute the values into the formula to get \(I = 0.1867 kg*m^2 + 0.56 kg*(0.3 m)^2\).
04

Calculate the final value

Calculate the final value to find the rotational inertia about the new axis. So, \(I = 0.1867 kg*m^2 + 0.56 kg*(0.3 m)^2 = 0.1867 kg*m^2 + 0.0504 kg*m^2 = 0.2371 kg*m^2\)

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Most popular questions from this chapter

A small lead sphere of mass \(25 \mathrm{~g}\) is attached to the origin by a thin rod of length \(74 \mathrm{~cm}\) and negligible mass. The rod pivots about the \(z\) axis in the \(x y\) plane. A constant force of \(22 \mathrm{~N}\) in the \(y\) direction acts on the sphere. (a) Considering the sphere to be a particle, what is the rotational inertia about the origin? (b) If the rod makes an angle of \(40^{\circ}\) with the positive \(x\) axis, find the angular acceleration of the rod.

Two identical blocks, each of mass \(M\), are connected by a light string over a frictionless pulley of radius \(R\) and rotational inertia \(I\) (Fig. 9-55). The string does not slip on the pulley, and it is not known whether or not there is friction between the plane and the sliding block. When this system is released, it is found that the pulley turns through an angle \(\theta\) in time \(t\) and the acceleration of the blocks is constant. (a) What is the angular acceleration of the pulley? ( \(b\) ) What is the acceleration of the two blocks? ( \(c\) ) What are the tensions in the upper and lower sections of the string? All answers are to be expressed in terms of \(M, I, R, \theta, g\), and \(t\).

Each of three helicopter rotor blades shown in Fig. 9-42 is \(5.20 \mathrm{~m}\) long and has a mass of \(240 \mathrm{~kg}\). The rotor is rotating at 350 rev/min. What is the rotational inertia of the rotor assembly about the axis of rotation? (Each blade can be considered a thin rod.)

A yo-yo (see Sample Problem \(9-13\) ) has a rotational inertia of \(950 \mathrm{~g} \cdot \mathrm{cm}^{2}\) and a mass of \(120 \mathrm{~g}\). Its axle radius is \(3.20 \mathrm{~mm}\) and its string is \(134 \mathrm{~cm}\) long. The yo-yo rolls from rest down to the end of the string. ( \(a\) ) What is its acceleration? (b) How long does it take to reach the end of the string? \((c)\) If the yo-yo "sleeps" at the bottom of the string in pure rotary motion, what is its angular speed, in rev/s? (d) Repeat \((c)\), but this time assume that the yo-yo was thrown down with an initial speed of \(1.30 \mathrm{~m} / \mathrm{s}\)

(a) Show that a solid cylinder of mass \(M\) and radius \(R\) is equivalent to a thin hoop of mass \(M\) and radius \(R / \sqrt{2}\), for rotation about a central axis. ( \(b\) ) The radial distance from a given axis at which the mass of a body could be concentrated without altering the rotational inertia of the body about that axis is called the radius of gyration. Let \(k\) represent the radius of gyration and show that $$ k=\sqrt{I / M} $$ This gives the radius of the "equivalent hoop" in the general case.
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