Physics for Scientists and Engineers
Physics for Scientists and Engineers
6th Edition
ISBN: 9781429281843
Author: Tipler
Publisher: MAC HIGHER
bartleby

Concept explainers

bartleby

Videos

Question
Book Icon
Chapter 9, Problem 82P

(a)

To determine

To Calculate: The moment of inertia of combination of platform-drum.

(a)

Expert Solution
Check Mark

Answer to Problem 82P

The moment of inertia of combination of platform-drum is 1.2 kgm2 .

Explanation of Solution

Given data:

Radius of the concentric drum, R=10 cm

  =(10 cm)( 1 m 100 cm)=0.100 m

Mass of the hanging object, M=2.50 kg

Distance through which the object falls, D=1.80 m

Time, t1=4.2s

Formula used:

Torque, τ=Iα

Where, I is the moment of inertia and α is the angular acceleration.

From Newton’s second law of motion:

  Fi=miai

Where, F represents force, m represents mass and a represents acceleration.

Second equation of motion is:

  s=v0t+12at2

Where, s is the displacement, t is the time, a is the acceleration and v0 is the initial velocity.

Calculation:

Apply the Newton’s second law of motion to the platform

  τ=TR=I0α.......(1)

Apply the Newton’s second law of motion to the weight:

  Fy=MgT=Ma.......(2)

Relation between angular acceleration and acceleration is α=aR

Substituting for α in equation (1) then the tension in the string is

  T=I0aR2

Now, substituting for the tension in equation (2) , and solve for the moment of inertia,

  MgI0aR2=MaI0=MR2( ga)a=MR2(ga1)..........(3)

The relation among the distance, acceleration and time is given by

  s=v0t+12at2

The intial velocity (v0) of the object is zero.

Substitute for the initial velocity (v0) of the object and distance in the above equation:

  D=12at2a=2Dt2

The moment of inertia of combinaiton of platform-drum is: I0=MR2( g t 2 2D1)=(2.50 kg)(0.100 m)2( ( 9.80 m/s 2 ) ( 4.20 s ) 2 2( 1.80 m )1)=1.18 kgm21.2 kgm2

Conclusion:

The moment of inertia of combination of platform-drum is 1.2 kg/m2 .

(b)

To determine

To Calculate: The total moment of inertia.

(b)

Expert Solution
Check Mark

Answer to Problem 82P

The total moment of inertia is 3.1 kgm2 .

Explanation of Solution

Given data:

Radius of the concentric drum, R=10 cm

  =(10 cm)( 1 m 100 cm)=0.100 m

Mass of the hanging object, M=2.50 kg

Distance fall by the object, D=1.80 m

Time, t2=6.8s

Formula used:

From the previous part:

  Itotal=MR2(gt22D1)

Calculation:

Substitute the values and solve for total moment of inertia:

  Itotal=MR2( g t 2 2D1)=(2.50 kg)(0.100 m)2( ( 9.80 m/s 2 ) ( 6.80 s ) 2 2( 1.80 m )1)=3.1 kgm2

Conclusion:

Total moment of inertia is 3.1 kgm2 .

(c)

To determine

To Calculate: The moment of inertia of the object.

(c)

Expert Solution
Check Mark

Answer to Problem 82P

The moment of inertia of the object is 1.9 kgm2 .

Explanation of Solution

Given data:

The total moment of inertia is 3.1 kgm2 .

The moment of inertia of combination of platform-drum is 1.2 kgm2 .

Formula used:

The moment of inertia of the object can be calculated by using the formula:

  I=ItotalI0

Calculation:

Moment of inertia of the object, I=ItotalI0

  =3.12kgm21.18 kgm2=1.9kgm2

Conclusion:

Moment of inertia of the object is 1.9 kgm2 .

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!
Students have asked these similar questions
CH 57. A 190-g block is launched by compressing a spring of constant k = = 200 N/m by 15 cm. The spring is mounted horizontally, and the surface directly under it is frictionless. But beyond the equilibrium position of the spring end, the surface has frictional coefficient μ = 0.27. This frictional surface extends 85 cm, fol- lowed by a frictionless curved rise, as shown in Fig. 7.21. After it's launched, where does the block finally come to rest? Measure from the left end of the frictional zone. Frictionless μ = 0.27 Frictionless FIGURE 7.21 Problem 57
3. (a) Show that the CM of a uniform thin rod of length L and mass M is at its center (b) Determine the CM of the rod assuming its linear mass density 1 (its mass per unit length) varies linearly from λ = λ at the left end to double that 0 value, λ = 2λ, at the right end. y 0 ·x- dx dm=λdx x +
Shrinking Loop. A circular loop of flexible iron wire has an initial circumference of 161 cm , but its circumference is decreasing at a constant rate of 15.0 cm/s due to a tangential pull on the wire. The loop is in a constant uniform magnetic field of magnitude 1.00 T , which is oriented perpendicular to the plane of the loop. Assume that you are facing the loop and that the magnetic field points into the loop. Find the magnitude of the emf E induced in the loop after exactly time 9.00 s has passed since the circumference of the loop started to decrease. please show all steps

Chapter 9 Solutions

Physics for Scientists and Engineers

Ch. 9 - Prob. 11PCh. 9 - Prob. 12PCh. 9 - Prob. 13PCh. 9 - Prob. 14PCh. 9 - Prob. 15PCh. 9 - Prob. 16PCh. 9 - Prob. 17PCh. 9 - Prob. 18PCh. 9 - Prob. 19PCh. 9 - Prob. 20PCh. 9 - Prob. 21PCh. 9 - Prob. 22PCh. 9 - Prob. 23PCh. 9 - Prob. 24PCh. 9 - Prob. 25PCh. 9 - Prob. 26PCh. 9 - Prob. 27PCh. 9 - Prob. 28PCh. 9 - Prob. 29PCh. 9 - Prob. 30PCh. 9 - Prob. 31PCh. 9 - Prob. 32PCh. 9 - Prob. 33PCh. 9 - Prob. 34PCh. 9 - Prob. 35PCh. 9 - Prob. 36PCh. 9 - Prob. 37PCh. 9 - Prob. 38PCh. 9 - Prob. 39PCh. 9 - Prob. 40PCh. 9 - Prob. 41PCh. 9 - Prob. 42PCh. 9 - Prob. 43PCh. 9 - Prob. 44PCh. 9 - Prob. 45PCh. 9 - Prob. 46PCh. 9 - Prob. 47PCh. 9 - Prob. 48PCh. 9 - Prob. 49PCh. 9 - Prob. 50PCh. 9 - Prob. 51PCh. 9 - Prob. 52PCh. 9 - Prob. 53PCh. 9 - Prob. 54PCh. 9 - Prob. 55PCh. 9 - Prob. 56PCh. 9 - Prob. 57PCh. 9 - Prob. 58PCh. 9 - Prob. 59PCh. 9 - Prob. 60PCh. 9 - Prob. 61PCh. 9 - Prob. 62PCh. 9 - Prob. 63PCh. 9 - Prob. 64PCh. 9 - Prob. 65PCh. 9 - Prob. 66PCh. 9 - Prob. 67PCh. 9 - Prob. 68PCh. 9 - Prob. 69PCh. 9 - Prob. 70PCh. 9 - Prob. 71PCh. 9 - Prob. 72PCh. 9 - Prob. 73PCh. 9 - Prob. 74PCh. 9 - Prob. 75PCh. 9 - Prob. 76PCh. 9 - Prob. 77PCh. 9 - Prob. 78PCh. 9 - Prob. 79PCh. 9 - Prob. 80PCh. 9 - Prob. 81PCh. 9 - Prob. 82PCh. 9 - Prob. 83PCh. 9 - Prob. 84PCh. 9 - Prob. 85PCh. 9 - Prob. 86PCh. 9 - Prob. 87PCh. 9 - Prob. 88PCh. 9 - Prob. 89PCh. 9 - Prob. 90PCh. 9 - Prob. 91PCh. 9 - Prob. 92PCh. 9 - Prob. 93PCh. 9 - Prob. 94PCh. 9 - Prob. 95PCh. 9 - Prob. 96PCh. 9 - Prob. 97PCh. 9 - Prob. 98PCh. 9 - Prob. 99PCh. 9 - Prob. 100PCh. 9 - Prob. 101PCh. 9 - Prob. 102PCh. 9 - Prob. 103PCh. 9 - Prob. 104PCh. 9 - Prob. 105PCh. 9 - Prob. 106PCh. 9 - Prob. 107PCh. 9 - Prob. 108PCh. 9 - Prob. 109PCh. 9 - Prob. 110PCh. 9 - Prob. 111PCh. 9 - Prob. 112PCh. 9 - Prob. 113PCh. 9 - Prob. 114PCh. 9 - Prob. 115PCh. 9 - Prob. 116PCh. 9 - Prob. 117PCh. 9 - Prob. 118PCh. 9 - Prob. 119PCh. 9 - Prob. 120PCh. 9 - Prob. 121PCh. 9 - Prob. 122PCh. 9 - Prob. 123PCh. 9 - Prob. 124PCh. 9 - Prob. 126PCh. 9 - Prob. 127PCh. 9 - Prob. 128PCh. 9 - Prob. 129P
Knowledge Booster
Background pattern image
Physics
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, physics and related others by exploring similar questions and additional content below.
Similar questions
SEE MORE QUESTIONS
Recommended textbooks for you
Text book image
Physics for Scientists and Engineers: Foundations...
Physics
ISBN:9781133939146
Author:Katz, Debora M.
Publisher:Cengage Learning
Text book image
College Physics
Physics
ISBN:9781938168000
Author:Paul Peter Urone, Roger Hinrichs
Publisher:OpenStax College
Text book image
College Physics
Physics
ISBN:9781285737027
Author:Raymond A. Serway, Chris Vuille
Publisher:Cengage Learning
Text book image
Principles of Physics: A Calculus-Based Text
Physics
ISBN:9781133104261
Author:Raymond A. Serway, John W. Jewett
Publisher:Cengage Learning
Text book image
University Physics Volume 1
Physics
ISBN:9781938168277
Author:William Moebs, Samuel J. Ling, Jeff Sanny
Publisher:OpenStax - Rice University
Text book image
Physics for Scientists and Engineers, Technology ...
Physics
ISBN:9781305116399
Author:Raymond A. Serway, John W. Jewett
Publisher:Cengage Learning
Moment of Inertia; Author: Physics with Professor Matt Anderson;https://www.youtube.com/watch?v=ZrGhUTeIlWs;License: Standard Youtube License