Go Back   Physics Help Forum > College/University Physics Help > Advanced Mechanics

Advanced Mechanics Advanced Mechanics Physics Help Forum

Reply
 
LinkBack Thread Tools Display Modes
Old Apr 14th 2009, 12:00 PM   #1
C.E
Member
 
Join Date: Apr 2009
Posts: 71
Angular rotation question

Hi, I am massively stuck on the following question and am examined on this stuff soon, could someone please help me?

A thin, uniform 15kg post of length 1.75 m is held vertically using a cable attached to the top of the post. A string attached to a 5kg mass passes over a smooth, massles pulley and is attached to the post 0.5m from the top. The post has a pivot at the bottom of it (i.e. that is what it rests on). Suddenly the cable snaps.

a. Find the angular acceleration of the post (with respect to the pivot) just after the cable snaps.
b.Will the angular acceleration from a be constant before the post hits the pulley? Why?
c.What is the acceleration of the 5kg mass the instant the pulley breaks? Is this constant? Why?

According to the back of my book the answers I should get are:
a. 2.65 rad/s^2
b. no, no explanation why given.
c. 3.31 m/s^2, n (no explanation given)

My attempt.
a. Torque (T)= moment of inertia (I) x angular acceleration (A)
T= 5g(1.75-0.5)=61.25
I= (MR^2)/4 = 15.3125
a=T/I = 4
(I don't see where I am going wrong for this part, any ideas?)
b. I think no because as the stick falls the line of action of the applied force changes and hence so does the torque. Is this explanation correct?
c. I though this acceleration was just g. As for whether it is constant I thought yes, but the answer is no.
C.E is offline   Reply With Quote
Old Apr 14th 2009, 12:23 PM   #2
Physics Team
 
Join Date: Feb 2009
Location: India
Posts: 365
(a) The moment of inertia of the rod about one of its end is ML^2/3 and not ML^2/4.

Further you have to take in to concederation the moment of inertia of the 5 kg mass as well.

HEnce the moment of inertia of the post and the 5 kg mass system will be

(ML^2)/3 + 5(1.75-0.5)^2 = 15*1.75*1.75/3 + 5*1.25*1.25 = 23.125 kgm^2

a = T/I = 61.25/23.125=2.65

( I could not visualise the problem clearly. Not even sure what I have done above is correct)

Last edited by Parvez; Apr 14th 2009 at 12:27 PM.
Parvez is offline   Reply With Quote
Old Apr 14th 2009, 12:29 PM   #3
C.E
Member
 
Join Date: Apr 2009
Posts: 71
Sorry, I did actually use /3 not /4 but have copied down what I did incorectly. Thanks for the help, I still have a few questions regarding part a though. In cases like these do you always have to find the moment of inertia of the entire system? I thought we were only bothered about the post for part a. The 5kg mass is not directly above the pivot so why is its moment of inertia 5(1.75-0.5)^2. Why are we able to treat the mass as a particle (i.e assume its contribution to the moment of inertia is mr^2)?
C.E is offline   Reply With Quote
Old Apr 14th 2009, 12:34 PM   #4
Physics Team
 
Join Date: Feb 2009
Location: India
Posts: 365
As I menssioned above I could not visualise the problem clearly. I diagram would be of much help.
Parvez is offline   Reply With Quote
Old Apr 14th 2009, 12:45 PM   #5
Physics Team
 
Join Date: Feb 2009
Location: India
Posts: 365
The moment cable breaks, the angular acclns of the post and that of the pully will be same.
Now linear accln a' = ra ( r= distance of 5 kg mass from the pivot)
=1.25*2.65 = 3.31 m/s/s

( just shooting in the dark !)

Last edited by Parvez; Apr 14th 2009 at 12:49 PM.
Parvez is offline   Reply With Quote
Old Apr 14th 2009, 12:47 PM   #6
Physics Team
 
Join Date: Feb 2009
Location: India
Posts: 365
Sorry. I dont know how to read what you have written.
Parvez is offline   Reply With Quote
Old Apr 14th 2009, 01:35 PM   #7
C.E
Member
 
Join Date: Apr 2009
Posts: 71
Ok, here is a diagram this should help. Why would the initial acceleration= initial angular acceleration?
Attached Thumbnails
Angular rotation question-pulley-diagram-2.jpg  
C.E is offline   Reply With Quote
Old Apr 15th 2009, 10:28 AM   #8
Physics Team
 
Join Date: Feb 2009
Location: India
Posts: 365
Refering to this diagram now every thing seems done wrong by me.
Was this diagram given with the problem in the book or you created it using your imagination ?

Last edited by Parvez; Apr 15th 2009 at 10:50 AM.
Parvez is offline   Reply With Quote
Old Apr 16th 2009, 05:42 AM   #9
C.E
Member
 
Join Date: Apr 2009
Posts: 71
I have literally copied the diagram out of the book. Do you know how to do it now? (It seems odd you got the correct answers by the wrong method)

Last edited by C.E; Apr 16th 2009 at 05:48 AM.
C.E is offline   Reply With Quote
Reply

  Physics Help Forum > College/University Physics Help > Advanced Mechanics

Tags
angular, question, rotation



Search tags for this page
Click on a term to search for related topics.
Thread Tools
Display Modes


Similar Physics Forum Discussions
Thread Thread Starter Forum Replies Last Post
Rigid body rotation question (Is my book's answers wrong?) s3a Kinematics and Dynamics 0 Jun 29th 2011 08:59 PM
Angular Frequency Question help need sonog4 Periodic and Circular Motion 1 May 4th 2011 05:09 AM
Urgent- a rotation question C.E Advanced Mechanics 3 Apr 19th 2009 11:06 PM
Angular Displacement Question!!! arminogue Kinematics and Dynamics 1 Apr 18th 2009 09:29 PM
angular velocity, center of rotation disclaimer Advanced Mechanics 0 Nov 3rd 2008 12:32 PM


Facebook Twitter Google+ RSS Feed