The questions we solve are too analogous..half the questions in mechanics can be solved using analogies..calling these analogies 'tricks' would be a crime! they are observations made by painfully analysing a set of problems based on a single concept...
Let me show you something:
Posted alongside are three different scenarios..encountered in different contexts. All lead to the same answer. Why should they not???
Let us denote the angle or repose of the incline in each case by 'R'..
in each case the block is to be kept at rest wrt. the incline. i want you to solve for each case:
1)the min
2)the max
3)and the optimum (when no friction acts)
value of the physical qty. specified (under the constraint of no relative motion b/w the block and the incline)..and try to observe.
Obviously i have left some thinking to you..this will take some effort!!
take you time..anyways, i won't be online for a week. next post next Sunday!
(in the third diagram the accn. of the wedge is maintained const. at 'a')
Yes, Anuj
ReplyDeleteThey lead to the same answers because if we analyse from the reference frame attached to the wedge, the force-body diagrams will be the same.
Let Y be the angle on the incline everywhere
First, the optimum case:
All 3 cases have mg acting downwards and N acting perpendicular to the wedge.
In all cases NcosY will be equal to mg
and NsinY will be equal to Mv^2/r, Mw^2r and MA respectively.
We can solve and get and obviously, V^2/r = W^2r = A
Since the angle of repose of the incline in each case is R, the coefficient of static friction is tanR. (is the angle of repose defined when the wedge is at rest?)
For the minimum case, friction should be taken to be acting is such a way that its component along the horizontal is in the opposite direction to NsinY.
The equations then will be:
Mg = NcosY + FsinY
NsinY - FcosY = Mv^2/r = Mw^2/r = MA
F in this case will be equal to tanRmg/(cosY+tanRsinY)
For the maximum case, friction will be taken in a direction so as to assist NsinY
The equations will reduce to
Mg + FsinY = NcosY
NsinY + FcosY = Mv^2/r = MA = MW^2r
and the value of F will be
tanRmg/(cosY - tanRsiny)
we can solve and get the required physical quantities in all 3 cases
what you have to get is a crisp, easy-to-remember expression! the questions are easy..but the extract is powerful..
ReplyDeleteyea, so how about the general expression
ReplyDeleteA = g(tan(theta) +(-) tanR) / (1 -(+) tanRtan(theta))
and simply set tanR = 0 for no friction case?
Again, I realise I may be completely missing your point.
In case, this does happen to be what you were asking for, do you recommend using this as a standard result?
you are nearly there...maybe you should now apply a bit of trigo???
ReplyDeletesure my friend,
ReplyDeleteA = G*tan(theta+(-)R)
plus for maximum
minus for minimum
and take R = 0 for no friction because if there is no friction, the angle of repose is 0.
Again, and more importantly, is is useful to remember this result? I mean, does it save time during JEE?
I am getting v=sqrt[rg{(tan(theta)-nu)}/{1+nu*tan(theta)}]
ReplyDeleteto find omega put v in w=v/r
to find acc. put a=v^2/r
if friction is acting max= when it is nu
min= put in place of nu as -nu
if no friction put nu=0
if angle of repose is given put nu=tan(repose)
which would come
v=sqrt[rg(tan{(theta)-(repose)}]
SIR PLEASE SOLVE THE PROBLEM GIVEN BELOW PLEASE!! A child of mass m sits in a swing of negligible mass suspended by a rope of length l. Assume that the dimensions of the child are negligible compared with 1.
ReplyDeleteHis father pulls the child back until the rope makes an angle of one radian with vertical, then pushes with a force F=mg along the arc of the circle releasing at the vertical :
a) How high up will the swing go?
b) How long did the father push?
ans:(a) theta=63degree above horizon (b) t=1.52(l/g)^1/2
for (i)simply apply the concept
ReplyDeletework done=change in potential energy
(ii)that's the question of SHM as it is the amplitude therfore its T/4
T=time period