Products
Rewards 
from HOLOOLY

We are determined to provide the latest solutions related to all subjects FREE of charge!

Please sign up to our reward program to support us in return and take advantage of the incredible listed offers.

Enjoy Limited offers, deals & Discounts by signing up to Holooly Rewards Program

HOLOOLY 
BUSINESS MANAGER

Advertise your business, and reach millions of students around the world.

HOLOOLY 
TABLES

All the data tables that you may search for.

HOLOOLY 
ARABIA

For Arabic Users, find a teacher/tutor in your City or country in the Middle East.

HOLOOLY 
TEXTBOOKS

Find the Source, Textbook, Solution Manual that you are looking for in 1 click.

HOLOOLY 
HELP DESK

Need Help? We got you covered.

Chapter 2

Q. 2.6

A bar is hanging in equilibrium in the position shown in Figure 2.14(a). Determine the potential energy of the bar in terms of \theta the counterclockwise angular position of the bar from its equilibrium position when (a) the datum is taken to be the horizontal plane at the bottom of the bar when in equilibrium, (b) the datum is taken as the horizontal plane through the mass center when the bar is in equilibrium, and (c) the datum is taken to be the horizontal plane through the pin support.

Step-by-Step

Verified Solution

(a) As the bar swings through an angle \theta , as illustrated in Figure 2.14(b), the mass center is a distance

h=\frac{L}{2} +\frac{L}{2} \left(1-\cos \theta \right)          (a)

and has a potential energy with respect to the datum of

V=mg\frac{L}{2} \left(2-\cos \theta \right)           (b)

(b) Using a horizontal plane through G as a datum, we have

V=mg\frac{L}{2} \left(1-\cos \theta \right)          (c)

(c) Using a horizontal plane through \omicron as a datum, we have

V= -mg\frac{L}{2}\cos \theta          (d)