/*! 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}

91Ó°ÊÓ

A disk of mass 3 kgand radius0.15 mhangs in the xy plane from a horizontal low-friction axle. The axle is 0.09 m from the center of the disk. What is the frequencyof small-angle oscillations of the disk? What is the period?

Short Answer

Expert verified

The frequency of small angle oscillation of the disk is 1.07Hz.

Step by step solution

01

Define the frequency of Oscillations.

The frequency of the oscillation is the number of oscillations in one second. If the particle completes one oscillation in T seconds, then the number of oscillations in one second or frequency is given as,

f=1T

The frequency of oscillation is measured in Hertz.

02

About the period of oscillations:

The period of the oscillations of a disk in a vertical plane through the point S is,

T=2πIsMgh ….. (1)

Here, Is is moment of inertia of the disk relative to Mis mass of the disk, his distance from the centre of mass of disk to the suspension point S and g is acceleration due to gravity.

03

Figure shows that the oscillations of a disk in a vertical plane: 

The diagram shows the oscillations of a disk in a vertical plane supported by a horizontal axis through a point S.

04

Find the frequency of the disk: 

The moment of inertia of the disk relative to pointOis,

IO=12MR2

Here, M is mass of the disk and is radius of the disk.

Thus, the moment of inertia of the disk relative to is,

Is=IO+Mh2=12MR2+Mh2

Substitute for into equation (1).

T=2Ï€12MR2+Mh2Mgh=2Ï€R2+2h22gh

Substitute 0.15 m for R, 0.09 forh , and 9.8m/s2 for g in the above equation.

T=2×3.140.15m2+20.09m2(9.8m/s2)0.09m

T=6.280.0225+0.0162m21.764m2/s2=6.280.148s=0.93s

Therefore, the period of the oscillations of a disk is 0.93s.

The frequency of small angle oscillation of the disk is,

f=1T

Substitute 0.93sfor T

f=10.93s=1.07Hz

Hence, the frequency of small angle oscillation of the disk is 1.07Hz.

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

According to the Bohr model of the hydrogen atom, what is the magnitude of the translational angular momentum of the electron (relative to the location of the proton) when the atom is in the 2nd excited state above the ground state(N=3)?

Redo the analysis, calculating torque and angular momentum relative to a fixed location in the ice anywhere underneath the string (similar to the analysis of the meter stick around one end). Show that the two analyses of the puck are consistent with each other.

As shown in figure, seven forces all with magnitude \(\left| {\overrightarrow F } \right| = 25{\rm{ N}}\) are applied to an irregularly shaped object. Each force is applied at a different location on the object, indicated by the tail of the arrow; the directions of the force differ. The distances shown in the diagram have these values: \(w = 9{\rm{ m}},{\rm{ }}h = 14{\rm{ m}}\)and\(d = 13{\rm{ m}}\). For each force, calculate the \(z\)-component of the torque due to that force, relative to location A (\(x\) to the right, \(y\)up, \(z\) out of the page). Make sure you give the correct sign. Relative to location A, what is the \(z\) component of the net torque acting this object?

Because the Earth is nearly perfectly spherical, gravitational forces act on it effectively through its center. Explain why the Earth’s axis points at the North star all year long. Also explain why the earth’s rotation speed stays the same throughout the year (one rotation per 24h). In your analysis, does it matter that the Earth is going around the sun?

In actual fact, the Earth is not perfectly spherical. It bulges out a bit at the equator, and tides tend to pile up water at one side of the ocean. As a result, there are small torque exerted on the Earth by other bodies, mainly the sun and the moon. Over many thousands of years there are changes in what portion of sky the Earth’s axis points towards (Change of direction of rotational angular momentum), and changes in the length of a day (change of magnitude of rotational angular momentum).

At a particular instant the location of an object relative to location \(A\) is given by the vector \({\overrightarrow r _A} = \left\langle {6,6,0} \right\rangle {\rm{m}}\). At this instant the momentum of the object is \(\overrightarrow p = \left\langle { - 11,13,0} \right\rangle {\rm{kg}} \cdot {\rm{m}}/{\rm{s}}.\) What is the angular momentum of the object about location \(A\)?

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.