1thnxx.....
yeah nw i cud slove d prob.....
bt u noe....still nt d (b) of ques 64 :(.....
mst b sum tiny miny error....
cud jus giv d equations....plz
hey cud u plzz slove ques. 63 part (c)
and ques 64....
of rotational meachnaics...of hc verma...pg 199.........
m sooo sry bt din feel lyk writing the ques as itz tooo big....
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5 Answers
62hmm...
See u have to do a very few things to solve such problems
1) Eliminate fear
2) Write down the equations
Dont be afraid to write what u think is right (Half the battle is in the mind :)
The equations come from 2-3 concepts only. There are not a 100 concepts floating... So these 3-4 equations (ofcourse i mean type) will solve all problems for u
1) Energy equation
2) Linear Momentum
3) Angular Momentum
In these questions... just give it one more shot:
a) Take All intitial conditions (just before collision)
b) Assume conditions after collision (ex: vx,vy, ω etc...
c) Apply Conservation of Angular Momentum if no external torque!
d) Apply Linear Momentum Cons. (No external force) {Even if there is ext torque or force, in collision problems, we ignore them bcos they act for "0" time!}
e) Energy balance if possible
This should suffice... pls try once mroe... let me know if u could.. other wise u will get the soln as soon as u ask :)
1
1Hope you got prob no.63.
Now, for prob.64,
(a) angular momentum jst after collision:= ang momentum due to the particle P= mgl(2gh)^0.5
find out angular speed from the equation:
mgl = 0.5 Iw^2
here, I=2m(l/2)^2 + m(l/2)^2
(b) find min value of “h†from the equation:
mgh= 0.5Iw^2 + mgl (put value of w got from previous equation)
(enrgy conservation.. wat happens is… energy gained from the striking particle is lost to gravitational potential energy 2mgl-mgl=mgl and rotational K.E. 0.5Iw^2)