Q. Show dat if a rod held at an angle theta to the horizontal and released, its lower end will not slip if the friction coefficient b/w the rod and the ground is greater than

nice rotational problem yaar.........

neways.........

the free body diagram is as posted below.......let the mass be 'm' the length be 'l'........let 'N' be the normal force which the rod experiences from the bottom..............." " be the coeff of friction..............

analysing rotational motion abt the pt rotation....ie the pt of contact with ground............

mgcos *l/2 =ml²/3 * ...........................1)     Torque abt pt P

so we have =3gcos /2l.......................2)

analysing accleration of centre of mass in horizontal direction.....

ma_x=N..............3)

analysing acceleration in vertical dirn....

mg-N= ma_y........................4)

but we know that ay is nothing but component of accn of c.m in vertical dirn........and as no slipping.....

accn of c.m= *l/2........

thus a_x=*l/2cos................................5) 

and a_y=*l/2sin ..................................6)

so using 1, 2, 3, 4, 5 and 6 and solving for

we get....=

hence proved.......

waise ek bande ne answer de diya hai but i will post my own solution it was a 6 out of 10 problem normally found in resonance paper.......

a=centre of mass accel.

A=angular acc.

Mg-N=ML/2Acos@

f=ML/2Asin@

torque about the end point

MgL/2cos@=IA

A=3gcos@/2L

applying SIR ISAAC NEWTON'S 2nd. LAW

friction=MLsin@3gcos@/4

uN=3Mgsin@cos@/4

Mg-N=MLcos@3gsin@/4l

N=Mg-3MGcos^@/4

u(Mg-3Mgcos^@/4)=3Mgsin@cos@/4

u=3sin@cos@/(4-3cos^@)

u=3sin@cos@/(1+3sin^@)