Friday, August 10, 2007

Irodov Problem 1.74

Let the force of friction between the bead and the rope be f, the tension in the string be T and let the accelerations of the masses M and m be and respectively directed towards the floor (downwards as shown in the picture).

Forces acting on M: There are two forces acting M, i) the tension in the string T and the force of gravity pulling it downwards Mg. The mass accelerates at a rate downwards, hence its dynamics can be written as,



Force acting on the bead : There are two forces acting on the bead, i) the force of friction f opposing the downward motion of the bead as it slides along the rope and ii) the force of gravity pulling it downwards mg. Thus, it dynamics are given by,




Relating the force of friction to the tension in the string : The missing piece is the relation between the force of friction and the tension in the string. For this let us consider the small section of the string that holds the bead. as shown in the figure. There are two forces acting on this section of the string, i) the tension T from the rest of the string pulling the section up and ii) the force of friction f from the bead pulling it down (as it opposes motion of the string relative to the bead and the string is moving up relative to the bead). The mass of the section of string is 0 and its acceleration is upwards. Thus, we have,




From (3) and (1) we can solve as follows,








Now the relative acceleration with which the bead moves up with respect to the rod is thus we have,





From (5) and (6) we have,

7 comments:

Anonymous said...

AMAZING i love this blog!!!!

But i have a doubt, doesn't Tension also act on the bead?

Krishna Kant Chintalapudi said...

Tension is present in the string around the pulley - it is like the tension in an elastic band. If you pull an elastic band with your hands then, the tension in the band pulls your hands together.

If you do not hold the elastic band tight and let the elastic band slide off, then the tension in the elastic string does not pull your hands together.

Tension will act on bodies rigidly connected to the string (there should be no relative motion between the thread and the connected body). The bead is not rigidly connected to the string, it is sliding off - hence tension does not act on it.

Hope this explanation helped.

Anonymous said...

thanks :)

Anonymous said...

he body is rigidly attached to the striso u mean to say that in this case the tension is coming in the string because of the frictional force and not by mg force

or should i conclude that the tension acts only when the body iss attached rigidly to the string

Krishna Kant Chintalapudi said...

If the string is not attached to the bead then the string cannot pull the bead i.e. the tension in the string cannot pull the bead. For example in order to pull a friend of yours towards you, you need to hold him tightly, slipping will not help. If there were no friction then the body would simply slide off.

Anonymous said...

gr8

Maharshi Roy said...

Amazing thinking you've got,,did you complete Irodov at class 11 itself,,cause I'm also in 11th and very much dedicated to the book,,,In fact,,I have tried to solve this question around 20 times,,but finally saw your solution,,it's amazing..