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AravindG
 4 years ago
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
AravindG
 4 years ago
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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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i 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

JamesJ
 4 years ago
Best ResponseYou've already chosen the best response.2No, 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 WorkEnergy 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.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0james why arent u bringing tension oor gravity into the poicture?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0James 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 \[\frac{3}{4}Mv^2\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0@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.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0when we equate the velocity and acceleration*

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0And I also had yet another typo, sorry.... \[\omega^2 = \frac{v^2}{r^2} \text{ not } v^2r^2\] :)

AravindG
 4 years ago
Best ResponseYou've already chosen the best response.0u see in eqns of moment of inertia like mr^2,mr^2/2

AravindG
 4 years ago
Best ResponseYou've already chosen the best response.0what is the r basically??

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0r 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.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0total energy is k.e+p.e right so y arent u guys including that anywhere ?

AravindG
 4 years ago
Best ResponseYou've already chosen the best response.0i rotate about axis q then r is??

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0@Aravind that's a more complicated moment of inertia, so it's not as simple to write down.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0change in kinetic energy=work done by net force=m*a

AravindG
 4 years ago
Best ResponseYou've already chosen the best response.0hey actually i did that qn

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0no. m*a is not the work done. It is the force.

AravindG
 4 years ago
Best ResponseYou've already chosen the best response.0pls tell the sides are made of rods

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0in ther question they have asked for a in terms of M,m.g but we have to include displacement here

AravindG
 4 years ago
Best ResponseYou've already chosen the best response.0wat is momentr of inertia?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0You have to do an integral to find that. Post a new question about calculating moment of inertia if you'd like to know how.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0And no we don't, not for the acceleration.

AravindG
 4 years ago
Best ResponseYou've already chosen the best response.0hey but salini gave me the answr 3mr^2

AravindG
 4 years ago
Best ResponseYou've already chosen the best response.0hey we did that remember?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0is athat an equilateral triangle?

AravindG
 4 years ago
Best ResponseYou've already chosen the best response.0made of uniform rods of length l

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0then use trignometry to find aq

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0arayila enniku orakam verunu anywaY PARA NOOKATE

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Using 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)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0siddhantsharan, why is Tr = Mr^2 / 2 * a / 2r that is I*a/2r why? and how?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Salini, 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. :)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1327926218152:dw@ siddhantsharan your 3rd step: acce of sphere = a (COM) + (angular Acceleration) / r

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Sorry. Typo error. Answer is still the same.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0YOu got the concept right?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Tr = Mr^2 / 2 * a / 2r here, is a/2r the angular acceleration of cylinder if so T=M*a*r/4 not Ma/4

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Wait I'll draw it for clarity.dw:1327878291024:dw

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0last doubt.....how did u get T*r ?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw: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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