Physics Help Forum Angular rotation question

 Apr 14th 2009, 12:00 PM #1 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.
 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.
 Apr 14th 2009, 12:29 PM #3 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)?
 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.
 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.
 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.
 Apr 14th 2009, 01:35 PM #7 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
 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.
 Apr 16th 2009, 05:42 AM #9 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.

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# A thin, uniform, 15.3 kg post, 2.11 m long, is held vertically using a cable and is attached to a 5.00 kg mass and a pivot at its bottom end (as shown below). The string attached to the 5.00 kg mass passes over a massless, frictionless pulley and pulls p

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