| Message |
|
|
sin3 + cos3
= (sin + cos)(sin2+cos2-sincos)
=(sin + cos)(1-(sin2)/2)
|
|
|
|
sin  + cos
= 2 [sin / 2+ cos/ 2]
= 2sin( +  /4]
This is the most common simplification ... though many more are possible
|
|
|
|
Delightful solution pcdagr
|
|
|
|
Pure rolling can be studied as a combination of translational motion and rotational motion about CM.
In the translational part:
The velocity of each point of the wheel is same as the CM. So the lowermost point has a speed v rightwards.
In the rotational part,
Each point on the rim has a velocity  r directed along the tangent. So the lowermost point has a velocity  r directed leftward (Assuming clockwise rotation)
So the net velocity of the lowermost point is the sum of the forward velocity v and the backward velocity r . If v = r then the net velocity of the point of contact is v - r = 0.
For the topmost point, the velocities due to translation and rotation are in the forward directions. So they add up to v + r. If v = r then topmost point moves with speed 2v.
|
|
|
|
I agree with arnie. As the velocity of rolling is constant, its acceleration is towards centre.
|
|
|
|
Let the equation be
a0 + a1x +a2x2 + .... +anxn = 0
Let the roots be x1, x2, .... xn
 xi = -an-1/ an
 xixj = an-2/ an
xixjxk = -an-3/ an
.
.
.
xix2 ...... xn = (-1)na0/ an
As the roots are x1, x2, .... xn
The equation can be written as
an(x-x1)(x - x2) ....(x - xn) = a0 + a1x +a2x2 + .... +anxn
Expand the LHS and compare. You'll end up with the relations given
|
|
|
|
Thanx for correcting me.
r = 20cm, the length of the horizontal section
On calculating, I am getting 3.5 2 = 4.949
|
|
|
|
There are two walls right?? ... One at the point where the block starts?? Assuming it is so ...
Acceleration of the block = qE/m
Time taken for first collision = (2dm/qE)
Time period, T = 2 (2dm/qE)
But the motion isn't SHM as there is no force directed towards an equilibrium point.
|
|
|
|
In a river boat problem, the boat tries to move perpendicular to the stream current. that analogy isn't true here.
|
|
|
|
Any system of mass creates a potential not only outside itself but throughout itself as well. Self PE corresponds to the part of the potential that it creates through itself. For the derivation of the formula see "Fundamentals of Mechanics" by IE Irodov
|
|
|
|
Please read "Fundamentals of Mechanics" by IE Irodov. Its a very good book on Mechanics ... but let me warn you ... its mathematical level is a bit higher than +2 level. Before reading that book... get an elementary idea abt vector calculas. That should seeyou through.
|
|
|
|
Let at any instant, the velocity vector of B makes an angle with the direction of motion of A.
Then, the velocity of approach
-(dl/dt) = u - u cos
-d 0 dl = u0T (1- cos)dt
d = uT -u 0 T cos dt ------------(1)
After sufficient time B moves on the same line as B, only separated by a distance, say x.
uT -u 0T cos dt = x
u 0T cos dt = uT - x ---------------(2)
From (1) and (2)
x=d
|
|
|
|
Let the area of cross section be A
Consider a fluid element in the horizontal portion, ata distance x from the axis of rotation and of length dx. Then centrifugal force on the element is (  Adx) 2 x.
Total centrifugal force on the horizontal fluid = 0r  A2 xdx. = A2 r2/2
where r= 20 cm
Now as the water rises to h (=5 cm), there is pressure imbalance on the horizontal section. This counterbalances the centrifugal force
A2 r2/2 = ghA
= (1/r) 2gh
|
|
|
|
@phyana
The acceleration of the elevator isn't g-a. Its a.
You are trying to work in the reference frame of earth. So the equation will be
Displacement of the coin wrt earth = Disp of the coin wrt lift + displacement of the lift wrt earth
-(1/2)gT2 = -h -(1/2)aT2
See, understand this, no matter which frame u work in, the equation will come out to be the same if the argument is sound
|
|
|
|
Ultimate solution Bipin. Excellent work
|
|
|
|
