Physics For Scientists And Engineers With Modern Physics, 9th Edition, The Ohio State University
Physics For Scientists And Engineers With Modern Physics, 9th Edition, The Ohio State University
9th Edition
ISBN: 9781305372337
Author: Raymond A. Serway | John W. Jewett
Publisher: Cengage Learning
bartleby

Concept explainers

bartleby

Videos

Question
Book Icon
Chapter 15, Problem 59AP

(a)

To determine

Tension in the rod at the pivot point.

(a)

Expert Solution
Check Mark

Answer to Problem 59AP

Tension in the rod at the pivot point is 2Mg_.

Explanation of Solution

Given that the mass of rod and ball is equal to M.

The tension at the pivot point is equal to the sum of the force of gravity on the rod and ball.

Write the expression for the force of gravity on the rod.

    Fr=Mg                                                                                                                     (I)

Here, Fr is the gravitational force on the rod, M is the mass of the rod, and g is the acceleration due to gravity.

Write the expression for the force of gravity on the ball.

    Fb=Mg                                                                                                                   (II)

Here, Fb is the force of gravity on the ball, and M is the mass of ball.

Write the expression for the tension in the rod.

    F=Fr+Fb                                                                                                             (III)

Here, F is the tension in the rod.

Conclusion:

Use expressions (I) and (II) in expression (III) to find F.

    F=Mg+Mg=2Mg

Therefore, tension in the rod at the pivot point is 2Mg_.

(b)

To determine

The tension in the rod at the point P when the system of rod and ball is at rest.

(b)

Expert Solution
Check Mark

Answer to Problem 59AP

The tension in the rod at the point P when the system of rod and ball is at rest is Mg(1+yL)_.

Explanation of Solution

At point P the whole weight of rod is not to be considered. At this point only yL portion of the rod only contributes. Therefore the tension at point P is equal to the sum of weight of rod due to yL portion of rod and the weight of the ball.

Force of gravity on the ball is equal to Mg. Write the expression for the force of gravity on yL portion of rod.

    Fp=Mg(yL)                                                                                                          (IV)

Write the expression for the tension.

    F=Fp+Fb                                                                                                              (V)

Conclusion:

Substitute Mg for Fb, and Mg(yL) for Fp in equation (V) to find F.

    F=Mg(yL)+Mg=Mg(1+yL)

Therefore, the tension in the rod at the point P when the system of rod and ball is at rest is Mg(1+yL)_.

(c)

To determine

The period of oscillation for small displacements.

(c)

Expert Solution
Check Mark

Answer to Problem 59AP

The period of oscillation for small displacements is 4π32Lg_.

Explanation of Solution

Write the expression for moment of inertia of the ball with respect to the point of pivot.

    Iball=ML2                                                                                                               (VI)

Here, Iball is the moment of inertia of the ball, M is the mass of the ball, and L is the distance between ball and pivot.

Write the expression for the moment of inertia of the rod rotating about one end.

    Irod=13ML2                                                                                                           (VII)

Here, Irod is the moment of inertia of the rod rotating about one end.

Write the expression for the total moment of inertia of system.

    I=Irod+Iball                                                                                                         (VIII)

Use expression (VI) and (VII) in expression (VIII) to find I.

    I=13ML2+ML2=43ML2                                                                                                     (IX)

Write the expression for the time period of the physical pendulum.

    T=2πImgd                                                                                                          (X)

Here, T is the time period of the physical pendulum, I is the moment of inertia of the pendulum, m is the mass of pendulum, and d is the distance from pivot to center of mass of rod and ball combination.

Write the expression to find the distance from pivot to center of mass of rod and ball combination.

    d=m1d1+m2d2m1+m2                                                                                                     (XI)

Here, m1 is the mass of rod, m2 is the mass of ball, d1 is the distance from pivot to center of mass of rod, and d2 is the distance between pivot to ball.

Conclusion:

Substitute M for m1 and m2, L2 for d1, and L for d2 in expression (XI) to find d.

    d=M(L2)+MLM+M=3L4

Substitute 43ML2 for I, 2M for m, and 3L4 for d in expression (X) to find T.

    T=2π43ML2(2M)g(3L4)=4π32Lg

Therefore, the period of oscillation for small displacements is 4π32Lg_.

(d)

To determine

The time period of oscillation if the length of the pendulum is 2.00m.

(d)

Expert Solution
Check Mark

Answer to Problem 59AP

The time period of oscillation if the length of the pendulum is 2.00m is 2.68s_.

Explanation of Solution

Write the expression for time period of oscillation of the pendulum as found in part (c).

    T=4π32Lg                                                                                                          (XII)

Conclusion:

Substitute 2.00m for L, and 9.80m/s2 for g in equation (XII) to find T.

    T=4π32(2.00m)(9.80m/s2)=2.68s

Therefore, the time period of oscillation if the length of the pendulum is 2.00m is 2.68s_.

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!

Chapter 15 Solutions

Physics For Scientists And Engineers With Modern Physics, 9th Edition, The Ohio State University

Ch. 15 - Prob. 5OQCh. 15 - Prob. 6OQCh. 15 - Prob. 7OQCh. 15 - Prob. 8OQCh. 15 - Prob. 9OQCh. 15 - Prob. 10OQCh. 15 - Prob. 11OQCh. 15 - Prob. 12OQCh. 15 - Prob. 13OQCh. 15 - Prob. 14OQCh. 15 - Prob. 15OQCh. 15 - Prob. 16OQCh. 15 - Prob. 17OQCh. 15 - Prob. 1CQCh. 15 - Prob. 2CQCh. 15 - Prob. 3CQCh. 15 - Prob. 4CQCh. 15 - Prob. 5CQCh. 15 - Prob. 6CQCh. 15 - Prob. 7CQCh. 15 - Prob. 8CQCh. 15 - Prob. 9CQCh. 15 - Prob. 10CQCh. 15 - Prob. 11CQCh. 15 - Prob. 12CQCh. 15 - Prob. 13CQCh. 15 - A 0.60-kg block attached to a spring with force...Ch. 15 - Prob. 2PCh. 15 - Prob. 3PCh. 15 - Prob. 4PCh. 15 - The position of a particle is given by the...Ch. 15 - A piston in a gasoline engine is in simple...Ch. 15 - Prob. 7PCh. 15 - Prob. 8PCh. 15 - Prob. 9PCh. 15 - Prob. 10PCh. 15 - Prob. 11PCh. 15 - Prob. 12PCh. 15 - Review. A particle moves along the x axis. It is...Ch. 15 - Prob. 14PCh. 15 - A particle moving along the x axis in simple...Ch. 15 - The initial position, velocity, and acceleration...Ch. 15 - Prob. 17PCh. 15 - Prob. 18PCh. 15 - Prob. 19PCh. 15 - You attach an object to the bottom end of a...Ch. 15 - Prob. 21PCh. 15 - Prob. 22PCh. 15 - Prob. 23PCh. 15 - Prob. 24PCh. 15 - Prob. 25PCh. 15 - Prob. 26PCh. 15 - Prob. 27PCh. 15 - Prob. 28PCh. 15 - A simple harmonic oscillator of amplitude A has a...Ch. 15 - Review. A 65.0-kg bungee jumper steps off a bridge...Ch. 15 - Review. A 0.250-kg block resting on a...Ch. 15 - Prob. 32PCh. 15 - Prob. 33PCh. 15 - A seconds pendulum is one that moves through its...Ch. 15 - A simple pendulum makes 120 complete oscillations...Ch. 15 - A particle of mass m slides without friction...Ch. 15 - A physical pendulum in the form of a planar object...Ch. 15 - Prob. 38PCh. 15 - Prob. 39PCh. 15 - Consider the physical pendulum of Figure 15.16....Ch. 15 - Prob. 41PCh. 15 - Prob. 42PCh. 15 - Prob. 43PCh. 15 - Prob. 44PCh. 15 - A watch balance wheel (Fig. P15.25) has a period...Ch. 15 - Prob. 46PCh. 15 - Prob. 47PCh. 15 - Show that the time rate of change of mechanical...Ch. 15 - Show that Equation 15.32 is a solution of Equation...Ch. 15 - Prob. 50PCh. 15 - Prob. 51PCh. 15 - Prob. 52PCh. 15 - Prob. 53PCh. 15 - Considering an undamped, forced oscillator (b =...Ch. 15 - Prob. 55PCh. 15 - Prob. 56APCh. 15 - An object of mass m moves in simple harmonic...Ch. 15 - Prob. 58APCh. 15 - Prob. 59APCh. 15 - Prob. 60APCh. 15 - Four people, each with a mass of 72.4 kg, are in a...Ch. 15 - Prob. 62APCh. 15 - Prob. 63APCh. 15 - An object attached to a spring vibrates with...Ch. 15 - Prob. 65APCh. 15 - Prob. 66APCh. 15 - A pendulum of length L and mass M has a spring of...Ch. 15 - A block of mass m is connected to two springs of...Ch. 15 - Prob. 69APCh. 15 - Prob. 70APCh. 15 - Review. A particle of mass 4.00 kg is attached to...Ch. 15 - Prob. 72APCh. 15 - Prob. 73APCh. 15 - Prob. 74APCh. 15 - Prob. 75APCh. 15 - Review. A light balloon filled with helium of...Ch. 15 - Prob. 78APCh. 15 - A particle with a mass of 0.500 kg is attached to...Ch. 15 - Prob. 80APCh. 15 - Review. A lobstermans buoy is a solid wooden...Ch. 15 - Prob. 82APCh. 15 - Prob. 83APCh. 15 - A smaller disk of radius r and mass m is attached...Ch. 15 - Prob. 85CPCh. 15 - Prob. 86CPCh. 15 - Prob. 87CPCh. 15 - Prob. 88CPCh. 15 - A light, cubical container of volume a3 is...
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
Principles of Physics: A Calculus-Based Text
Physics
ISBN:9781133104261
Author:Raymond A. Serway, John W. Jewett
Publisher:Cengage Learning
Text book image
Physics for Scientists and Engineers
Physics
ISBN:9781337553278
Author:Raymond A. Serway, John W. Jewett
Publisher:Cengage Learning
Text book image
Physics for Scientists and Engineers with Modern ...
Physics
ISBN:9781337553292
Author:Raymond A. Serway, John W. Jewett
Publisher:Cengage Learning
Text book image
Physics for Scientists and Engineers: Foundations...
Physics
ISBN:9781133939146
Author:Katz, Debora M.
Publisher:Cengage Learning
Text book image
Classical Dynamics of Particles and Systems
Physics
ISBN:9780534408961
Author:Stephen T. Thornton, Jerry B. Marion
Publisher:Cengage Learning
Text book image
Physics for Scientists and Engineers, Technology ...
Physics
ISBN:9781305116399
Author:Raymond A. Serway, John W. Jewett
Publisher:Cengage Learning
SIMPLE HARMONIC MOTION (Physics Animation); Author: EarthPen;https://www.youtube.com/watch?v=XjkUcJkGd3Y;License: Standard YouTube License, CC-BY