So equating torque is completely out of use as angular acceleration would be different as R decreases right? But isn't the ribbon pulled in a way that the ribbon has a constant linear velocity so the ribbon in contact with disc will also have constant velocity. Assuming no slipping between...
Homework Statement
A disc is free to rotate about an axis passing through its center and perpendicular to its plane. The moment of inertia of the disc about its center is I. A light ribbon is tightly wrapped over it in multiple layers. The end of the ribbon is pulled out at a constant velocity...
Ok so I picked the the equation balancing the forces pulling it up and down the incline.
Hence f + m(1-n)gSinQ = T
So tension here would be equal to weight of the part hanging so T = mng
f=m(1-n)gCosQ u
So on substituting these values in the above equation, I got;
u= [n-(1-n)SinQ]/(1-n)CosQ...
So I take the initial part of chain on incline as a system and draw it's free body diagram. The exernal forceks are normal by surface, gravitational force, friction down the incline and tension up the incline by the other part of chain. But I am finding difficulty finding friction coefficient...
There is a frictionless pulley at the tip of incline. The dimensions of the pulley is to be neglected.(view image)
At the start.
The equation I get is W(by friction) = (mlgCosQ(1-n)^2 u)/2
Mg(h+H/2)*
Which is like all the water is concentrated at the center of cylinder (by height) and finding work done to carry that to a height h. I hope this is correct
Yeah sorry I meant to type Adx/dt = AH/T
dx/dt = H/T
If I integrate dx from 0 to x and dt from 0 to t, I get
x= H t /T
dW/dT = M(dx/dt)g (h+x)
P= M H/T g (h+ Ht/T)
The answer is a function of a general time 't' so maybe this should be correct, thanks a lot for your support sir.
Am I allowed to...
Homework Statement
A uniform chain of mass 'm' and length 'l' rests on a rough incline (inclination is angle 'Q') with its part hanging vertically. The chain (inclined) starts moving up the incline (and the vertical part moving down) provided the hanging (vertical) part equals to 'n' times...
Density = M/A H (A is base area of cylinder
So M/AH = dm/A dx
So dm = Mdx/H
As water is decreasing at a uniform rate, Adx/dt = constant
dW/dt =(Mdx/Hdt)h'g
Even if I write Adx/dt = H/T, I don't get the answer as a function of time.
Homework Statement
A completely filled cylindrical tank of height H contains water of mass M. At a height h above the top of the tank there is another wide container. The entire water from the tank is to be transferred into the container in time T such that level of water in tank decreases at a...