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Question

A solid cylinder of mass m=4 kg and radius R=10 cm has a string wrapped around it as shown in figure.The cylinder is held horizontal by fixing the two free ends of the string to the hook on the ceiling such that both the strings are exactly vertical. The cylinder is released to fall under gravity. Find the linear acceleration (a) of the cylinder.


A
2g3
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B
g3
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C
g2
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D
g
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Solution

The correct option is A 2g3
Applying Newton's 2nd law in direction of acceleration on the cylinder,


mg2T=ma ...(i)
For rotational motion of the cylinder, equation of torque
τcom=Icomα
2TR=(mR22)α ...(ii)
τmg=0 about centre.

For string to not slip, tangential acceleration should be equal to acceleration of centre of mass a
at=αR
a=αR ...(iii)
Substituting in Eq. (ii),
T=ma4
Substituting in Eq. (i),
mgma2=ma
or 3ma2=mg
a=2g3
Here, a is the linear acceleration of the centre of mass of the cylinder.

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