/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Q9Q Figure 10 - 26 shows a uniform ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Figure 10 - 26shows a uniform metal plate that had been square before 25 %of it was snipped off. Three lettered points are indicated. Rank them according to the rotational inertia of the plate around a perpendicular axis through them, greatest first.

Short Answer

Expert verified

The rank of the points having a maximum moment of inertia isc>a>b

Step by step solution

01

Step 1: Given data

The figure of 25 % snipped off square-shaped uniform metal plate.

02

Understanding the concept

We can rank the points according to the moment of inertia using the relation between the moment of inertia and the radius of gyration of the elements.

Formulae are as follows:

I=mh2

Where, h =Radius of gyration, mis mass, I is the moment of inertia.

03

Determining the rank of the points having a maximum moment of inertia

Here,

I=mh2

So, the final moment of inertia will be maximum for the axis which consists of elements having more h (which are more distant from the axis).

From this, it can be concluded that the moment of inertia about an axis through point c is maximum followed by (a) and then (b).

Rank is,c>a>b

Therefore, the points can be ranked according to the moment of inertia about the axis passing through it using its relationship with the radius of gyration.

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.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The uniform solid block in Fig 10-38has mass 0.172kg and edge lengths a = 3.5cm, b = 8.4cm, and c = 1.4cm. Calculate its rotational inertia about an axis through one corner and perpendicular to the large faces.

A gyroscope flywheel of radius 2.83 cmis accelerated from rest at14.2rads3 until its angular speed is 2760revmin.

(a) What is the tangential acceleration of a point on the rim of the flywheel during this spin-up process?

(b) What is the radial acceleration of this point when the flywheel is spinning at full speed?

(c) Through what distance does a point on the rim move during the spin-up?

A force is applied to the rim of a disk that can rotate like a merry-go-round, so as to change its angular velocity. Its initial and final angular velocities, respectively, for four situations are: (a) -2°ù²¹»å/²õ, 5°ù²¹»å/²õ ; (b)2°ù²¹»å/²õ, 5°ù²¹»å/²õ ; (c)-2°ù²¹»å/²õ, -5°ù²¹»å/²õ ; and (d)2°ù²¹»å/²õ, -5°ù²¹»å/²õ. Rank the situations according to the work done by the torque due to the force, greatest first.

In Fig. 10-50, two6.20kgblocks are connected by a mass-less string over a pulley of radius2.40cmand rotational inertia.7.40×10−4kg/m2The string does not slip on the pulley; it is not known whether there is friction between the table and the sliding block; the pulley’s axis is frictionless. When this system is released from rest, the pulley turns through0.130radin91.0msand the acceleration of the blocks is constant. What are (a) the magnitude of the pulley’s angular acceleration, (b) the magnitude of either block’s acceleration, (c) string tensionT1, and (d)T2string tension?

Figure 10 - 27shows three flat disks (of the same radius) that can rotate about their centers like merry-go-rounds. Each disk consists of the same two materials, one denser than the other (density is mass per unit volume). In disks 1and 3, the denser material forms the outer half of the disk area. In disk 2, it forms the inner half of the disk area. Forces with identical magnitudes are applied tangentially to the disk, either at the outer edge or at the interface of the two materials, as shown. Rank the disks according to (a) the torque about the disk center, (b) the rotational inertia about the disk center, and (c) the angular acceleration of the disk, greatest first.

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.