Chapter 1: Q29E (page 296)

Calculate the moment of inertia of each of the following uniform objects about the axes indicated. Consult Table 9.2 as needed.
(a) A thin 2.50-kg rod of length 75.0 cm, about an axis perpendicular to it and passing through (i) one end and (ii) its center, and (iii) about an axis parallel to the rod and passing through it.
(b) A 3.00-kg sphere 38.0 cm in diameter, about an axis through its center, if the sphere is (i) solid and (ii) a thin-walled hollow shell.
(c) An 8.00-kg cylinder, of length 19.5 cm and diameter 12.0 cm, about the central axis of the cylinder, if the cylinder is (i) thin-walled and hollow, and (ii) solid.

Short Answer

Expert verified

(a) For rod, we will use equations from table 9.2.
(b) Thus, the solid is .

(ii) Thus, the thin-walled is $0.072 k g . m^{2}$ .

Step by step solution

Step:-1 explanation

Mass of the rod = 2.5 kg

Length of the rod = 75 cm

$= \frac{75}{100} = 0.75 c m l = \frac{1}{3} M L^{2}$

Step:-2 solution

Put the value here,

$= \frac{1}{3} (2.5 \times 0.75) = 0.469 k m . m^{2}$

Mass of the rod = 2.5 kg

Length of the rod = 75 cm

$= \frac{75}{100} = 0.75 m l = \frac{1}{12} M L^{2}$

Putting the value here,

$= \frac{1}{12} (2.5 \times 0.75) = 0.117 k m . m^{2}$

For a very thin rod, since the masses are near to the axis, thus the moment of inertia =0

Step:-3 solid

given M =3.00 kg

$d = 38.0 c m r = \frac{d}{2} = \frac{38}{2} = 19 c m$

given M =3.00kg

d =38.0 cm

Step:-4 Concept

$\begin{aligned} I = \frac{1}{2} \times 3 \times (19)^{2} \\ = \frac{2}{5} \times 3 \times 361 \\ = 0.4 \times 3 \times 361 \\ = 433.2 \\ I = 0.043 kg \cdot m^{2} \end{aligned}$

Step:-5 thin-walled hollow shell

given M =3.00 kg

$d = 38.0 c m r = \frac{d}{2} = \frac{38}{2} = 19 c m$

Step:-6 calculation

$l = \frac{2}{3} M R^{2} = \frac{2}{3} M {(\frac{D}{2})}^{2} = 0.072 k g . m^{2}$

Step:-7 explanation

$d = 12 r = \frac{d}{2} r = 6$

Step:-8 Solution

$l = \frac{1}{2} M R^{2}$

Put the value here

$l = M {(\frac{D}{2})}^{2} = 0.0288 k g . m^{2}$

Thus, the answer $0.0288 k g . m^{2}$ .

Step 9:-explanation

$d = 12 r = \frac{d}{2} r = 6$

Step:-10 solution

$l = \frac{1}{2} M {(\frac{D}{2})}^{2} = \frac{1}{2} \times 8 \times 6^{2} = \frac{8 \times 36}{2} = 144 = 0.0144 k g . m^{2}$

Hence, the solution is $0.0144 k g . m^{2}$ .

Unlock Step-by-Step Solutions & Ace Your Exams!

Full Textbook Solutions
Get detailed explanations and key concepts
Unlimited Al creation
Al flashcards, explanations, exams and more...
Ads-free access
To over 500 millions flashcards
Money-back guarantee
We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Short Answer

Step by step solution

Step:-1 explanation

Step:-2 solution

Step:-3 solid

Step:-4 Concept

Step:-5 thin-walled hollow shell

Step:-6 calculation

Step:-7 explanation

Step:-8 Solution

Step 9:-explanation

Step:-10 solution

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Physics Textbooks

Space Physics

Work Energy and Power

Scientific Method Physics

Electrostatics

Thermodynamics

Fundamentals of Physics

Study anywhere. Anytime. Across all devices.