A uniform solid cylinder of mass M and radius | vedclass

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Question

Tension in the both string will be same.

$m g-T=m a....(1)$

also Inertia solid cylinder is $M R^{2} / 2$ Torque at centre of cylinder due to bot

Vedclass · Accepted Answer

$\frac{{4mg}}{{M + 4m}}$

Vedclass · Answer

Tension in the both string will be same.

$m g-T=m a....(1)$

also Inertia solid cylinder is $M R^{2} / 2$ Torque at centre of cylinder due to both string $2 T 	imes R=I \alpha=M R^{2} \alpha / 2 \ldots(2)$

also, $a=\alpha R....(3)$

From equations $(1),(2)$ and $(3)$

we get Tension $T=\frac{M m g}{M+4 m}$

A uniform solid cylinder of mass $M$ and radius $R$ rotates about a frictionless horizontal axle. Two similar masses suspended with the help two ropes wrapped around the cylinder. If the system is released from rest then the acceleration of each mass will be

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