What is the period of a book swinging like a pendulum?

AI Thread Summary
The discussion revolves around calculating the period of a book swinging like a pendulum, with specific dimensions of 28 cm and 19 cm. Participants explore various moments of inertia equations, including I=(ML^2)/12 and I=(ML^2)/3, but encounter incorrect results for the period. The correct approach involves using the parallel axis theorem to account for the book's rotation about a corner rather than its center. One participant suggests a period of 0.83 seconds using a simplified pendulum model, which raises concerns about the accuracy of this method. Ultimately, the conversation emphasizes the importance of correctly applying the principles of rotational inertia to achieve an accurate calculation of the pendulum's period.
jellyman
Messages
4
Reaction score
0
1. In the figure below, a book is suspended at one corner so that it can swing like a pendulum parallel to its plane. The edge lengths along the book face are 28 cm and 19 cm. If the angle through which it swings is only a few degrees, what is the period of the motion?

W0358-N.jpg


2. I=(ML^2)/12
I=(ML^2)/3
T=2∏√I/M*G*dcom


3. I got dcom by using the distance formula and it's .1692 m
Then I tried to using both inertia equations and using length, width. Then I plugged in all the numbers. (multiple attempts)

All wrong answers (.788s, .394s, 1.062s)






Appreciate your help! :)

.
 
Physics news on Phys.org
You don't have the right moments of inertia.
 
Oh.

Then that leaves me with I= M(a2 + b2)/12.

That gave me a period of .476 second which has been marked wrong.
 
Yep - that's the wrong moment of inertia as well.
That is for an oblong rotating about it's center.

You book is not rotating about it's center - otherwise it could not act as a pendulum.

Look up: parallel axis theorem.
 
debad27da7daea89152bd3c7b5d34dd7.png


I|| = \frac{1}{12}Mdcom2 + Mdcom2

This doesn't make sense though because it leaves me with an M on the top in the equation for T.EDIT: I asked somewhere else and they used T = 2π√[L/g] and got .83 seconds. This doesn't sound quite right. It's too easy this way!
 
Last edited:
$$I_{CM}=\frac{M}{12}(a^2+b^2)$$
##r## is the distance from the corner to the center of the book; by pythagoras: $$r^2=\frac{a^2}{4}+\frac{b^2}{4}$$... therefore, by the parallel axis theorem: $$I=I_{CM}+Mr^2=\cdots$$... you finish up.
 
EDIT: I asked somewhere else and they used T = 2π√[L/g] and got .83 seconds. This doesn't sound quite right. It's too easy this way!
That would be pretty normal ... he's modeled the book as a simple pendulum.

For a simple pendulum ##I=ML^2## ... completing the calculations above will tell you how taking the mass distribution into account affects the period.
http://en.wikipedia.org/wiki/Pendulum_(mathematics)#Compound_pendulum
 

Similar threads

Back
Top