/*! 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} Q3-3.4-15E A rotating flywheel is being tur... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A rotating flywheel is being turned by a motor that exerts a constant torque T (see Figure 3.10). A retarding torque due to friction is proportional to the angular velocity v. If the moment of inertia of the flywheel, is Iand its initial angular velocity is, find the equation for the angular velocity v as a function of time. [Hint:Use Newton’s second law for rotational motion, that is, moment of inertia * angular acceleration = sum of the torques.]

Short Answer

Expert verified

The equation of angular velocity is Ó¬(t)=TK+(Ó¬o-TK)e-KTI.

Step by step solution

01

Find the equation for the angular velocity

Here the notations are T= torque for motor, = angular velocity, I = moment of inertia and

Ó¬o= initial angular velocity.

Acc. To the question retarding torque due to friction is proportional to the angular velocity so, T1=-KÓ¬ (K is proportionality constant)

Now, moment of inertia × angular velocity = sum of the torques

IdÓ¬dt=T-KÓ¬IdÓ¬T-KÓ¬=dtVariable separatingITlnT-KÓ¬K=t+CIntegrating on both sidesITlnT-KÓ¬=-KtI+C1T-KÓ¬=Ce-Kt/IÓ¬(t)=T-Ce-Kt/IK

02

Find the value of C

Put Ó¬0=Ó¬0then C=T-KÓ¬o

Ó¬(t)=TK+(Ó¬o-TK)e-KTI

Hence, the equation of angular velocity is role="math" localid="1663966683149" Ó¬(t)=TK+(Ó¬o-TK)e-KTI.

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

Transmission Lines.In the study of the electric field that is induced by two nearby transmission lines, an equation of the formdzdx+g(x)z2=f(x)arises. Letf(x)=5x+2andg(x)=x2. If z(0)=1, use the fourth-order Runge–Kutta algorithm to approximate z(1). For a tolerance ofε=0.0001, use a stopping procedure based on the absolute error.

Use the fourth-order Runge–Kutta algorithm to approximate the solution to the initial value problem\({\bf{y' = ycosx,y(0) = 1}}\) , at \({\bf{x = \pi }}\). For a tolerance of \({\bf{\varepsilon = 0}}{\bf{.01}}\) use a stopping procedure based on the absolute error.

It was noon on a cold December day in Tampa: 16°C. Detective Taylor arrived at the crime scene to find the sergeant leaning over the body. The sergeant said there were several suspects. If they knew the exact time of death, then they could narrow the list. Detective Taylor took out a thermometer and measured the temperature of the body: 34.5°C. He then left for lunch. Upon returning at 1:00 p.m., he found the body temperature to be 33.7°C. When did the murder occur? [Hint: Normal body temperature is 37°C.]

A parachutist whose mass is 100 kgdrops from a helicopter hovering 3000 m above the ground and falls under the influence of gravity. Assume that the force due to air resistance is proportional to the velocity of the parachutist, with the proportionality constant b3=20 N-sec/mwhen the chute is closed andb4=100 N-sec/m when the chute is open. If the chute does not open until 30 sec after the parachutist leaves the helicopter, after how many seconds will he hit the ground? If the chute does not open until 1 min after he leaves the helicopter, after how many seconds will he hit the ground?

Falling Body.In Example 1 of Section 3.4, page 110, we modeled the velocity of a falling body by the initial value problem \({\bf{m}}\frac{{{\bf{dv}}}}{{{\bf{dt}}}}{\bf{ = mg - bv,v(0) = }}{{\bf{v}}_{\bf{o}}}{\bf{ = 0}}\)under the assumption that the force due to air resistance is –bv. However, in certain cases the force due to air resistance behaves more like\({\bf{ - b}}{{\bf{v}}^{\bf{r}}}\), where \({\bf{(r > 1)}}\) is some constant. This leads to the model \({\bf{m}}\frac{{{\bf{dv}}}}{{{\bf{dt}}}}{\bf{ = mg - b}}{{\bf{v}}^{\bf{r}}}{\bf{,v(0) = }}{{\bf{v}}_{\bf{o}}}\) (14).To study the effect of changing the parameter rin (14),take \({\bf{m = 1,}}\,\,{\bf{g = 9}}{\bf{.81,}}\,\,{\bf{b = 2}}\) and \({{\bf{v}}_{\bf{o}}}{\bf{ = 0}}\).Then use the improved Euler’s method subroutine with \({\bf{h = 0}}{\bf{.2}}\) to approximate the solution to (14) on the interval \(0 \le {\bf{t}} \le 5\)for \({\bf{r = 1}}{\bf{.0,}}\,\,{\bf{1}}{\bf{.5}}\) and 2.0. What is the relationship between these solutions and the constant solution\({\bf{v(t) = }}{\left( {\frac{{{\bf{9}}{\bf{.81}}}}{{\bf{2}}}} \right)^{\frac{{\bf{1}}}{{\bf{r}}}}}\)?

See all solutions

Recommended explanations on Math 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.