| Message |
|
|
Let the tension in the string of the bobbin A be T1 . The frictional force on the bobbin a is f, directed up the incline.
Then,
T1 + f - mAgsin = mAaA
T1R - fR = (IA - mAR2 )A
aA = AR
aA + AR = BR
Let the tension in the string emanating from the inner radius of the bobbin B be T2 .
T2R - T1R = IBB
Br = aC - CR
Let the tension in the string connecting the bobbin C and the mass m be T3 .
T3+mg - 2T2 = mCaC
T3r = (IC - mCR2)C
For the mass m,
mg - T3 = ma
a = aC + r
Please verify whether or not these equations are correct.
|
|
|
|
Got it. ....
See consider a length x of the rope from the end where F2 is acting. Let the tension at the other end be T.
Mass of this portion of the rope = (m/L)x
Acceleration of the rope = (F1 - F2)/m
So
T - F2 =[ (m/L)x] [(F1 - F2)/m]
T = F2 +[(F1 - F2)/L]x
|
|
|
|
I also don't think this is correct answer. Will reply soon. Can you provide the answer?? Maybe I can do some backtracking.
|
|
|
|
The car flies off tangentially from the outer circle. Let its speed be v at the moment it flies off the circular path. Then at that moment its at a height h above the ground and the base of the embarkment is at a distance hcosec along the incline.
When the car flies off the embarkment, it flies as a projectile projected horizontally from a height h. The quantity R is measured from the base of the embarkment and not from the point vertically below the point of projection.
Range of the car =  R2 - h2 cot2 
Then,
v (2h/g) =  (R2 - h2 cot2 )
v = g(R2 - h2 cot2 )/2h
|
|
|
|
The correct solution is below
|
|
|
|
Let at any moment the man be at a depth h below the horizontal. As both his friends exert equal forces, the strings make equal angles with the vertical. let at this instant the string makes an angle with the vertical.
The forces on the man are the two tensions T making an angle  with the vertical and his weight mg acting down. Its stated in the question that the friends are slowly pulling him out. So his acceleration in the vertical direction may be assumed to be zero.
Then,
2T cos = mg
T = mg / 2cos
As the man moves up increases. Hence cos decreases and so the value of T increases. So force exerted by the friends increases as the man comes up.
At a depth d,
T = mg / 2cos
Now, cos = d/(d/2)2 + d2 = 2/ 5
T = 5mg/4d
|
|
|
|
Let the velocity of the man wrt the river be u.
Then the net velocity of the man is the vector sum of his velocity wrt water and velocity of the water wrt earth.
Then as the man wants to swim perpendicular to the flow,
ucos = V
usin = v
Dividing the above equations
tan = v/V
= tan-1(v/V)
|
|
|
|
Let the convex mirror be on the left and the mirror on the right. Drop a perpendicular from the object on to the principal axis. Let the point of intersection of this perpendicular and the principal axis (say A) be at a distance x from the convex mirror. Let the point object be O.
We can imagine a linear object AO of height h (= 6 cm). The tip of the image of this linear object AO will give the position of the images of the point O.
For the convex mirror,
u = x
1/v + 1/u = -1/F
where F is the magnitude of f
v = -xF/(x+F)
Magnification, m = -v/u = h'/h
h' being the height of image and h being the height of the object.
h' = hF/(x+F)
h' is +ve, and hence the image is erect. Hence, the image of the point O by the convex mirror is at a height hF/(x+F)
For the plane mirror,
The image is formed at distance 40 - x behind the mirror.
Distance between the two images is measured without parallax means that its measured along the line joining the two images, ie along A'A''
A'A'' = O'O''2+ BA''2 = [hF/(x+F) + 40 + 40 - x]2 + [h - hF/(x+F)]2
[hF/(x+F) + 40 + 40 - x]2 + [h - hF/(x+F)]2 = 4 cm
But, there are two variables: x and F. For finding the focal length of the mirror, we must know x. Please check the question again
|
|
|
|
Let the refractive index of the lens be 
The focal length of any resulting optic device can be found by letting parallel rays from infinity (parallel to the principal axis) fall on the device and finding where the image is formed.
When the plane surface isn't silvered
When parallel rays fall on the curved surface,
u = -
Then by the refraction formula
 /v - 1/(-) = (-1)/R
v =R/(-1)
This serves as object for refraction at the plane surface.
v' = v / = R/(-1)
Hence focal length of this plano convex lens = R/(-1) = 40 cm
Once the plane surface is silvered
Now, when the parallel rays from infinity fall on the unsilvered surface,
u = -
Then by the refraction formula
/v - 1/(-) = (-1)/R
v =R/(-1)
This serves as the object for reflection from the silvered plane surface. For the reflection at this surface, the image is formed at the same distance but on the opposite side.
v' = -R/( -1)
This serves as the object for the next refraction at the unsilvered curved surface.
1/v'' - /v' = (1-)/R
v'' = -R/2(-1)
Hence, focal length of this new optical device = R/2(-1) = 20 cm.
|
|
|
|
|
|
|
Lets prove this by integration. Consider a thin circular area element of radius x inside the given circle and concentric with the given circle. The width of this element be dx.
If this element is straightened out, its approximately a rectangular strip of length
2 x and width dx.
Area of the strip dA = 2 x dx
A = 0 r 2x dx = r2 /2
|
|
|
|
Consider a particle colliding with a wall. You can resolve the approach velocity perpendicular to the wall and parallel to it. The component parallel to the wall remains the same even after the collision. The component perpendicular to the wall reverses its direction and its magnitude becomes e times the previous value.
Apply this analogy to all problems. If instead of a ball you have atwo balls, the component of approach velocity along the line joining them changes e times.
|
|
|
|
Let the equation of the position a particle undergoing SHMbe
x = A sin(2ft)
Velocity of the particle
v =2fA cos(2ft)
KE of the particle
K = (1/2)m[2fA cos(2ft)]2
K = (2f2A2/m)[1- cos(4ft)]
Frequency of K = 2f
|
|
|
|
