IIT JEE , AIEEE Forum - Physics , Mechanics

shreyasnivas

ok jaysonantony is right.. nobody has the patience to read all that..even tho it probably took u 1 hour..

i will ans the 4th q.
but try and imagine this.
firstly: you should know that the normal reaction does not always pass thru the centre of mass when body is under force.
please see the diagram i attached(its crude but u will get an idea)

in that example, the force is pushing right, so the body has a tendency to topple clockwise.. to cancel this, the normal reaction tries to produce an opposite torque, and it goes as far to the right as possible to do this.

notice that the value of normal reaction must be so as to cancel MG, so the only limit to the torque is the perpendicular distance from centre.

next:in an irregular body the normal reaction will again do the same job.i it will go as far as it can to nullify the external torque. but there is a limit here too. hence it will also topple after this limit.(there is no real difference)

next:the flaw in your argument is that you have forgotten torque.
think about it now..no the numerical value of normal reaction will not change so long as the MG stays same.

next:the point will stay locked as a hinge simply because the friction acting at that point is preventing translational motion!!!!
a concept about friction states that it is independent of surface area(dont ask why even i find it hard to accept!!), so the fristion will stop it moving, but friction does not atop rotation, whereas it plays a vital role in most cases(as jayson says it is not always needed, but in all rolling experiences involving a rotatong body on the ground it is always there)

now i know that i have not answered all your doubts, but seriously dude, nobody will read it.. im sorry. break it up and see the success you get!!

now please rate me i have worked very hard on this!! :)

shreyas

Ask & Discuss Questions with Community & Experts