1someone doing this .....?
Please tell me how to do this type of problems. Consider the situation shown. A small disc of mass 1 kg is released from the top of a fixed wedge which is inclied at 60° to the horizontal. The coefficient of friction of surface of wedge varies as u=5x, where x is the distance of point on surface from the top A of the wedge (g=10m/s2). Disc is released from the top of inclined. 
1) What is the frictional force acting on the disc when it covers a distance of 1 cm? 2) What is the distance covered after which the disc will start slipping? 3) What is the velocity of the centre of mass when it starts slipping? Ans 1) 5√3 N 2) 5√3 cm 3) 1 m/s Please tell me your working method......
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14 Answers
1can u please tell me what is coefficient of surface of wedge ??? What does it mean ?? Otherwise it's easy.
1yaar diya to hai coefficient of friction varies as u = 5/x
1!@ aveek or govind...plz do it yaar ....ccoz i am getting the opp answers .....for (1) am getting ans 2
and for (2) am getting 1
:(
1i missed coefficient "of friction" so edited it...yaar pranav post ur solution, i m much more interested in the solution... and the answer that i hv written is as it was given in the aakash solutions
11st answer : f is constant ... it does not depend on whether distance covered is 1 cm or 10000 cm.
mg cos60 - f = ma (translation)
f. R = I . (a/R) (torque)
I have taken pure rolling for this condition as in the 2nd part of the quest. it has been asked when the disc will start slipping. So obviously it's rolling in the initial case.
so you get f=0.5 ma
so a = g/√3
so f = 0.5 x (g/√3) = 5/√3
The next parts are easy after this. :-)
1i am not getting ,,,,,trabslation ..qn????
how mgcos@
??? but that downwards na/./