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

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- AravindG

|dw:1327757617642:dw|

- anonymous

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

- JamesJ

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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## More answers

- anonymous

james why arent u bringing tension oor gravity into the poicture?

- anonymous

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
\[\frac{3}{4}Mv^2\]

- anonymous

@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

when we equate the velocity and acceleration*

- AravindG

wndful

- anonymous

And I also had yet another typo, sorry....
\[\omega^2 = \frac{v^2}{r^2} \text{ not } v^2r^2\]
:)

- JamesJ

(@jem, thanks)

- AravindG

i hav a doubt

- AravindG

u see in eqns of moment of inertia like mr^2,mr^2/2

- AravindG

what is the r basically??

- anonymous

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.

- anonymous

total energy is k.e+p.e right so y arent u guys including that anywhere ?

- AravindG

|dw:1327769992982:dw|

- AravindG

|dw:1327770030252:dw|

- AravindG

i rotate about axis q then r is??

- AravindG

|dw:1327770065239:dw|

- anonymous

@Aravind that's a more complicated moment of inertia, so it's not as simple to write down.

- anonymous

change in kinetic energy=work done by net force=m*a

- AravindG

hey actually i did that qn

- anonymous

no. m*a is not the work done. It is the force.

- AravindG

pls tell the sides are made of rods

- anonymous

in ther question they have asked for a in terms of M,m.g but we have to include displacement here

- AravindG

wat is momentr of inertia?

- anonymous

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.

- anonymous

And no we don't, not for the acceleration.

- AravindG

hey but salini gave me the answr 3mr^2

- anonymous

when?

- AravindG

hey we did that remember?

- anonymous

is athat an equilateral triangle?

- AravindG

ys

- AravindG

made of uniform rods of length l

- AravindG

q is centre of mass

- anonymous

then use trignometry to find aq

- AravindG

axis is q

- AravindG

aq is l/root 3 then??

- anonymous

3*m*(aq)^2

- AravindG

that is 3 ml^2

- AravindG

srry ml^2

- anonymous

annu

- AravindG

u gud in maths?

- AravindG

salini?

- anonymous

arayila enniku orakam verunu anywaY PARA NOOKATE

- AravindG

http://www.twiddla.com/736142

- anonymous

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)

- anonymous

siddhantsharan, why is Tr = Mr^2 / 2 * a / 2r
that is I*a/2r why? and how?

- anonymous

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. :)

- anonymous

|dw:1327926218152:dw|@ siddhantsharan
your 3rd step: acce of sphere = a (COM) + (angular Acceleration) / r

- anonymous

Sorry. Typo error. Answer is still the same.

- anonymous

YOu got the concept right?

- anonymous

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

- anonymous

Wait I'll draw it for clarity.|dw:1327878291024:dw|

- anonymous

last doubt.....how did u get T*r ?

- anonymous

|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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