consider the arrangement shown in fig The string is wrapped around a uniform cylinder which rolls without slipping .The other end of the string is passed over a massless ,frictionless pulley to a falling weight .determine acceleration of falling mass m in terms of only the mass of the cylinder M ,the mass m and g

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consider the arrangement shown in fig The string is wrapped around a uniform cylinder which rolls without slipping .The other end of the string is passed over a massless ,frictionless pulley to a falling weight .determine acceleration of falling mass m in terms of only the mass of the cylinder M ,the mass m and g

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|dw:1327757617642:dw|
i consider that both of the object have the same mass. F = ma F = mg , (gravitational force which pulling the square object) F = (m+m)a ,(grouping thses 2 object into the system) F/2m = a
No, that analysis isn't right. The cylinder has a moment of inertia, \[ I = \frac{Mr^2}{2} \] where \( r \) is its radius. Hence if the cylinder has velocity \( v \), then because \( v = \omega r \) where \omega is the angular velocity. Thus the total kinetic energy is \[ KE_{linear} + KE_{rotational} = \frac{1}{2}Mv^2 + \frac{1}{2}I\omega^2 \] \[ = \frac{1}{2}Mv^2 + \frac{1}{2}Mr^2\frac{v^2}{r^2} \] \[ = Mv^2 \] Now, given that, you can use conservation of energy to find the energy of the cylinder when the weight \( m \) has fallen a distance \( h \). From that, use the Work-Energy Theorem and the fact \[ \Delta U = \Delta KE_{block} + \Delta PE_{m} = Work = \int F \cdot dy = \int ma \ dy \] to find the acceleration \( a \) on the block.

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james why arent u bringing tension oor gravity into the poicture?
James is correct, though I believe to correct a small typo, the rotational kinetic energy should be \[ \frac{1}{2} I \omega^2 = \frac{1}{2}\left(\frac{Mr^2}{2}\right)v^2r^2 = \frac{1}{4}Mv^2\] so the total kinetic energy is \[\frac{3}{4}Mv^2\]
@salini we automatically include tension when we equate the velocity of the center of mass of the cylinder with the velocity of the hanging weight, and the F in the integral happens to be the gravitational force.
when we equate the velocity and acceleration*
wndful
And I also had yet another typo, sorry.... \[\omega^2 = \frac{v^2}{r^2} \text{ not } v^2r^2\] :)
(@jem, thanks)
i hav a doubt
u see in eqns of moment of inertia like mr^2,mr^2/2
what is the r basically??
r is the radius of the cylinder. The moments of inertia of differently shaped objects are not the same. For a disc and a solid cylinder, I = mr^2/2 (assuming we're rotating about the axis poking through the radial center), whereas for a hoop or a hollow cylinder I = mr^2.
total energy is k.e+p.e right so y arent u guys including that anywhere ?
|dw:1327769992982:dw|
|dw:1327770030252:dw|
i rotate about axis q then r is??
|dw:1327770065239:dw|
@Aravind that's a more complicated moment of inertia, so it's not as simple to write down.
change in kinetic energy=work done by net force=m*a
hey actually i did that qn
no. m*a is not the work done. It is the force.
pls tell the sides are made of rods
in ther question they have asked for a in terms of M,m.g but we have to include displacement here
wat is momentr of inertia?
You have to do an integral to find that. Post a new question about calculating moment of inertia if you'd like to know how.
And no we don't, not for the acceleration.
hey but salini gave me the answr 3mr^2
when?
hey we did that remember?
is athat an equilateral triangle?
ys
made of uniform rods of length l
q is centre of mass
then use trignometry to find aq
axis is q
aq is l/root 3 then??
3*m*(aq)^2
that is 3 ml^2
srry ml^2
annu
u gud in maths?
salini?
arayila enniku orakam verunu anywaY PARA NOOKATE
http://www.twiddla.com/736142
Using Forces, Tr = Mr^2 / 2 * a / 2r ----- where a is acceleration of block which will be double of cylinder. And T is Tension. => T = Ma/4 Now plugging in equation for block, mg - T = ma mg - Ma/4 = ma Therefore, a = 4mg / ( M + 4m)
siddhantsharan, why is Tr = Mr^2 / 2 * a / 2r that is I*a/2r why? and how?
Salini, In the fig. the string is attached to top of the sphere. From rotation dynamics we know that acceleration of the point at the top of sphere = a (COM) + (angular Acceleration) / r = a + a ( pure rolling => a = Angular acceleration / r) = 2a. Where a is acceleration of COM of sphere which is what we consider the sphere's acceleration to be during Force Concept. Now along the string Acceleration has to be equal. Hence, a ( top of sphere) = a (block) Therefore a(block) = 2a. In the answer I have assumed a to be acceleration of block and solved for a. Therefore Acceleration of sphere = a /2 Cheers. :)
|dw:1327926218152:dw|@ siddhantsharan your 3rd step: acce of sphere = a (COM) + (angular Acceleration) / r
Sorry. Typo error. Answer is still the same.
YOu got the concept right?
Tr = Mr^2 / 2 * a / 2r here, is a/2r the angular acceleration of cylinder if so T=M*a*r/4 not Ma/4
Wait I'll draw it for clarity.|dw:1327878291024:dw|
last doubt.....how did u get T*r ?
|dw:1327880069640:dw| Hence , Force on sphere is Tension. Torque of Tension = r x F = r x T Hence Torque is Tr.

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