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A man pulls a solid cylinder (initially at rest) horizontally by a massless string as shown.The string is wrapped on the cylinder and the cylinder performs pure rolling (that is ,rolling without slipping).Mass of the cylinder is 100kg radius is pie metre and tension in string is 100N.Then the angular speed of the cylinder after one revolution will be

Asked by m.nilu 20th October 2018, 5:06 PM
Answered by Expert
Torque τ = I×α ......................(1)
where I moment of inertia = (1/2)M×R2 , where M is mass and R is radius of cyliner. α is angular acceleration
Torque = force × distance = 100 × π  Nm = (1/2)×100π2  × α or α = 2/π rad/s2
Angular speed ω is given by,  ω2 = ω02 +2× α × θ ,
where ω0 is initial angular speed (zero in this case) and θ is angular displacement.
Hence angular speed after one revolution is given by, ω2 =2 × (2/π)×2π = 8,   or ω = √8 rad/s
Answered by Expert 21st October 2018, 1:52 AM
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