tag:blogger.com,1999:blog-5997147895880585179.post8198259973634717626..comments2022-01-23T00:41:47.969+05:30Comments on Solutions to I E Irodov - Physical Fundamentals of Mechanics: Irodov Problem 1.283Unknownnoreply@blogger.comBlogger3125tag:blogger.com,1999:blog-5997147895880585179.post-36266830473959115822010-05-03T08:10:03.018+05:302010-05-03T08:10:03.018+05:30Torque is T = F X r, if r =0 then T = 0. So when ...Torque is T = F X r, if r =0 then T = 0. So when torque is considered about an axis, torque due to all forces that pass through the axis will have zero torque about this axis. In the solution the torque is evaluated about the axis passing through the point of contact. Hence all the forces at the point of contact do not exert any torque.Krishna Kant Chintalapudihttps://www.blogger.com/profile/07024636005277277405noreply@blogger.comtag:blogger.com,1999:blog-5997147895880585179.post-58565052485015862572010-04-21T12:11:08.701+05:302010-04-21T12:11:08.701+05:30Hi,
I have a doubt in problem of spinning top pre...Hi, <br />I have a doubt in problem of spinning top precession. The centre of mass of spinning top is now rotating in a circle of radius r’=l*sin(theta). <br />This means there has to be a Centripetal force mr’.w’.w’, this horizontal force is provided by point of contact. But this force will also have a torque on the top which <br />Should add up with torque due to its own weight… but this term does not appear in solution. May be we can neglect this term if w’ is too small. <br />But yes, the solution provided by you matches with the one in book :-)<br />I tried to search this on internet but dint find anything, can u send me ur e-mail id , mine is pradyumn@gmail.com, it will be faster to communicate that way <br /><br />Thanks <br />PradyumnaPradyumnanoreply@blogger.comtag:blogger.com,1999:blog-5997147895880585179.post-63215229386640739472009-03-12T11:52:00.001+05:302009-03-12T11:52:00.001+05:30This comment has been removed by a blog administrator.Anonymousnoreply@blogger.com