the moment of inertia of a solid flywheel of radius 0.1m about its axis is 0.1kgm2. a tangential force of 2 kg wt is applied through the circumference of flywheel with the help of string and mass M. the acceleration of the mass is

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Adikesh Kumar · Answer

ANS: 2m/s 2 equate the torque I alpha = force * r (r = radius of wheel) 0.1*a/r = 2*10* r ( a = accelaration) therefore a=2m/s 2 ( r = 0.1m)

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the moment of inertia of a solid flywheel of radius 0.1m about its axis is 0.1kgm2. a tangential force of 2 kg wt is applied through the circumference of flywheel with the help of string and mass M. the acceleration of the mass is

the moment of inertia of a solid flywheel of radius 0.1m about its axis is 0.1kgm2. a tangential force of 2 kg wt is applied through the circumference of flywheel with the help of string and mass M. the acceleration of the mass is

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the moment of inertia of a solid flywheel of radius 0.1m about its axis is 0.1kgm2. a tangential force of 2 kg wt is applied through the circumference of flywheel with the help of string and mass M. the acceleration of the mass is

the moment of inertia of a solid flywheel of radius 0.1m about its axis is 0.1kgm2. a tangential force of 2 kg wt is applied through the circumference of flywheel with the help of string and mass M. the acceleration of the mass is

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