Some of you were confused abt. velocity constraints..so, for a few days, lets restrict ourselves to rigid body constraints..
for practice and initiation, lets solve the following problems:
Given the velocity of one body, find that of the other. (don't blame me if you find these trivial)..
1.vy=vtanA
ReplyDelete2.V=vcosA
1. vy = vcotA
ReplyDelete2. v(of mass m)= vsecA
so, after all i was right in giving these easy problems..
ReplyDeletemessed up the tan's and sec's..
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ReplyDeleteidiotic mess up by me
ReplyDeletecorrect piyush..but you must get these right on the first attempt. no good on the second!
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ReplyDelete1.vy=vcota
ReplyDelete2. v=vseca
kya yaar..do minute pehle tak to thik tha??
ReplyDeleteI THINK PIYUSH NEW ANSWER IS CORRECT???I ALSO GET SAME
ReplyDeleteis my ans correct??(see the second post)
ReplyDeleteno and no..no-one has got it right.
ReplyDeletehey people..this is pathetic! these problems are real elementry..
ReplyDeletebhaiya "1.vcotA"
ReplyDeleteye kaise galat ho sakta hai
ye question to rotation ke subjective solved me hai
lets end this:
ReplyDelete1)Vy=VCotA
2)V(of mass m)=VCosA
ankush, you second part was wrong..
okk..
ReplyDeletebhaiya plz tell my mistake
let velocity of m be u,...and 2m velocity is given v
x1+x2= constant ......(x1 is distance frm pulley to m,..x2 is length of rope from pulley to 2m)
x1+x2 = constant
- dx1/dt = dx2 / dt
u= dx2/dt.....(i)
now,v=d(x2cosA) /dt
=>v=cosA dx2/dt
from (i) => v = u cosA
u= v sec A
while differentiating, you have assumed A to be unchanging with time..
ReplyDeletehe he, use this:
components of velocity of any 2 pts. on a tight string/rod along the dirn. of the string/rod must be equal..
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ReplyDeletegot my mistake....plz check this soln
ReplyDeletefrom the above soln
u= dx2/dt.....(i)
v=d(x2cosA) /dt
but we need dA /dt if cosA is not constant
here's the 3rd eqn
x2sinA =constant
distance between pulleys never changes
got dA/dt from this...used some trigo
finally got u = vcosA
how can we solve the question given in physics workshop(mechanics-2) qus -61,62,63 with help of COR theorem
ReplyDeletebhaiya 2nd question
ReplyDeleteVcos theta = 2 V(of m)
so V of m = V cos theta upon 2
nahi second mein string ke along velocity equate karni hai add nahi karenge dusra component dusri sting ke along hai
ReplyDeletePLEASE REPLY ANUJ SIR THE PROBLEM I ASKED.HERE IN YOUR PROBLEM YOUR ANSWER IS CORRECT...
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ReplyDeletebhaiya...some more of these....(a lil more difficult ones..) would be really helpful...!
ReplyDeletethnx a lot..!
sambhav can you understand my doubt
ReplyDeletehi amritpal..i have a solution for your problem...
ReplyDeletei have assumed a center of rotation for the rod, and the rest is fairly simple..
plz. you never want to assume a center of rotation in such problems, and without that the solution is complex. are you not satisfied with the problems i give??
plz. yaar, stick to what your teachers teach.
hi sambhav, thy will be done.
ReplyDeletePLEASE ANUJ SIR WHAT IS THE TOTAL KINETIC ENERGY OF THE LADDER WHEN THE LADDER TOP MAKES AN ANGLE THETA WITH THE WALL.PLEASE GIVE DETAIL BHAIYA.
ReplyDeletehi amritpal. the total KE is equal to the loss in PE.
ReplyDeletewhich is:
mg(L/2)(1-CosA)..if you assume A to be angle made by the rod with the vertical.