IIT JEE , AIEEE Forum - Physics , Mechanics

nishant_88

Please give the answer with working of HC Verma2 Pg.137 Q no.62, 63 and 64.

Rates assured for all proper answers.

shyam_nair_s

I can try the 62nd a) and c) part.

shyam_nair_s

here is the a part

LET B the topmost point of the incline.

Net force on the particle at A and B, F = mgsinA

Work done to reach from point B to C (horizontal position of the circular path on the other side) = mgh

= mgR (1-cosA )

So total work done = mg {lsinA + R (1-cosA)}

Net change in KE = work done

½ mv²= mg {lsinA+R (1-cosA)}

V = {2g[R (1-cosA) + lsinA]}^{1/2--------------------ANS}

^{hope u got it.}

shyam_nair_s

And the c part....

mv²/R= mgcosA

V ²= RgcosA-------------------(1)

Again ½ mv² = mg (R- RcosA)

V² = 2gr(1-cosA)----------------(2)

From 1 and 2,

CosA = 2/3

A= Cos inverse 2/3......................................ANS

where A is the angle which the radius makes with the vertical

best of luck...

nishant_88

guys please help